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Hi, everyone, welcome back.

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So in this video, we are going to discuss about discussion, distinct, no innuendo.

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So let's try to read the problem statement and let's try to analyze what is the problem and what can

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be the possible solution.

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The input will contain two things.

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First is basically, Neddie, and second thing, basically, Brindusa is.

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Second is basically windows size, so window size is three, that means consider three elements one,

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two and one.

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Now, what other windows can be possible?

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So remove one now and add this tree so another window can be two and three.

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Now, remove these two and add this four.

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So now the window will become one, three, four.

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So you can see window sizes three.

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So that's why I have three elements.

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Now, I will remove this element and I will add this element.

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So now the window is three for entry.

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So these are the possible windows.

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So these are all possible windows.

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Offices three, offices three now after figuring out all the possible windows what to do, you need

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to find out how many unique elements are there.

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So in this case, two unique elements.

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Are there one or two in this window?

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I have three unique elements.

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One to entry in this window.

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I have three unique elements, one, three and four.

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And in this window I have two unique elements.

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So what will be my output?

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My output will be the collection of all these.

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So my output will be two, three, three and two.

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So this will be my output for input one.

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So let's try to see what is the output for and what, when and where is the output.

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Yes, so this is two, three, three and two.

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And that is what we are done here, two, three, three and two.

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So I hope you have the question.

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If not, let's try again now.

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This time.

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Do with me rate.

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So what do you do?

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Window size is three.

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OK, so that means consider the first three elements.

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So consider the first three elements and this will be my window.

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Now what I will do.

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So remove this element and we need to maintain the size three.

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So I will add this element.

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Now, this is my next window.

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Which is two, one and three now, again, remove this element and take this element, so this will

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be my new window, one, three and four.

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Now, again, remove this element and consider this element, so this will be my new window, that will

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be three, four and three.

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So now I hope you know, what are all the possible windows of Sastry?

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And after figuring out all the windows, what do you need to find out the number of unique elements?

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So how many unique elements we have here to how many unique elements?

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Three.

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How many elements?

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Three, how many unique elements.

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So this will be my answer to three.

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Three, two.

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Now, let's talk about this example.

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So window size is one to consider, one element.

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Now, remove this element and add a new element.

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So, again, window size is one, so remove this element, add this element.

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So again, window size is one, so only one element.

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So remove this element.

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Add this element.

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So since the window size is one, so that's why.

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The number of elements in the window will be one and how many unique elements will be there since the

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window is containing only one element.

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So the number of unique element will always be one.

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So this will be my answer for this input.

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And we can see that the answer for our photo is one one, one one.

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So now you have understood the question properly.

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Now let's discuss how we can solve this question.

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So this is very simple and very easy.

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So what do will do?

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I will maintain one map.

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And this is my answer.

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Answer at a rate we need to return when Eddie, we can see we need to return one Eddie now Inmet But

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I will do the developer will be the key will be basically the element and the value will be the count

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of that element in window of size.

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W rate.

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Window size is three.

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So let's see how we can solve this problem.

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So what do you do?

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First of all, the window sizes, three cents, window sizes three to one.

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So what I will do, I will insert one in the map and each count this one.

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So first of all, I need to consider this window, right, because the window sizes three.

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So I am considering three elements.

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Now, the next is elemental.

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So insert element in the map.

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And what is the count?

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Count is when now the next element is when one is already present in the map.

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So we will update its count and now the count of one is two and that we will stop here because we need

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to consider only three elements.

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We will stop here and we will do so in MAP.

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How many elements are there?

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Two elements are present.

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Two entries are there in the map.

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So my answer will be to write.

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What is to do is basically map out size.

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The number of entries in the map.

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Fine.

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Very good.

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Now, let's start now what I need to do, I will consider one more element.

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So let me raise this first.

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So what do we do, we have considered these three elements.

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Now I will consider this element.

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So ultimately what I will do elementary is not present in the map, so I will insert and its count is

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one.

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So now the window size has become four, but window has to be of size three.

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So we will delete the last element.

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So what I will do, I will go to the map and what I will do, I will reduce its count by one.

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I am removing one right.

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I am removing one so each count will reduce by one.

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So this is the new count for one.

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And what need to do.

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We are considering this window.

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So how many elements are there?

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Again, my size and my map glasses will give you three because there are three elements.

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Now let's move forward again.

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So what I will do now, I will consider this element.

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So I will insert this element in the map and it's counted one, but now we have considered the four

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elements, but the window sizes three.

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So I will remove this element from the map.

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So remove two.

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So I'm removing two.

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So the count will decrease by one.

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So the count will become zero.

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And since the count is zero, that means we need to remove this entry from the map.

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We need to remove this entity from the map because the count becomes zero.

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And now for this window, how many elements are there?

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How many elements are there?

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That is my map.

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Size and map is containing three entries.

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So three.

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Now it will consider the last element registry, so I will insert old trees already present, so I will

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increase its count by one so the count becomes two and we need to remove element one.

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So I will go to eliminate one and I will reduce its count to zero.

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And since the count becomes zero, that means you need to remove this element from the map.

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You will remove this element from the map.

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And how many distinct elements are there.

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So my map daudzai, that will give you two because two elements are present in the map and that is your

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answer.

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You will reach here and it will stop.

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You are trading already at right.

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And you will reach the end of the area.

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So you will stop.

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And this is your answer to three, three, two, which is the right answer.

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Perfect.

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Now let's talk about this.

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So this is our map and let's quickly so we know size is one, so I am standing here, I will insert

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this element in the map.

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So the counties one and this is my answer, Victor.

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Answer.

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Eddie, what is my map that says it is containing one element?

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So that's why one map is containing one entity.

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So one next element is one.

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So I will increase its count.

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It's going to become two.

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But you need to remove this element so its count will decrease again.

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So the count again becomes one now for finding out the number of unique elements.

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The number of entries in the map is one.

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So that's why the output is one.

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Now, I will continue this element two, so I will insert two and the count is one.

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But I need to remove this element.

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So the count is one.

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I will decrease its count by one to the count become zero and zero means remove.

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So I will remove it.

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And the number of unique elements is basically the number of countries in the map and no fantasias,

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only one.

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So that's why one.

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Now, again, I will consider this element to so I will increase its count.

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Byron, so the count becomes two, but I need to remove this element so the government will become one

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again, and how many unique elements are there?

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So maybe containing only one, and that's why the output is one.

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And finally, you will reach the end of the year end.

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You will stop.

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And this is your answer.

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One one, one one.

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Which is correct.

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So I hope you have understood the question very well and the solution also very well.

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And the part is also going to be very, very simple.

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So what do you need to do?

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You need to maintain one map map of element and count.

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And what does discount discount this discount of the element in window size, the blue gentlemen in

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NSW and my map that says both sides will give you the number of unique elements, will give you the

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number of entries in the map and you need to maintain this on Saturday because we need to return one

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at a.

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Simple, so the cold is going to be very, very simple and you can all caused this problem very easily.

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So we'll be writing the code for this problem in the next video.

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And I will say you did.

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Thank you.