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Welcome back to the
continuation of the solution.

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We stopped at mean right here.

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And the last 2 parts that we've got
to do are display the smallest number

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and display the largest number.

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And then of course we have to handle
quit which is really easy to do.

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All right, so let me scroll
back down here and clean up a

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little bit of this white space.

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All right, so this is
where I right here.

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So now the user hasn't typed in a
p or an a or an m, maybe they've

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typed in an s, which is the smallest.

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So we'll do another
else if ladder here.

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We'll say else if the selection
is equal to a capital s or

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selection is a lowercase s.

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In this case, we need to calculate
the smallest number in the List.

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There's a lot of ways to do this.

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But again let's deal with the case
where the list is empty, right.

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So we'll say if numbers
just like we have before.

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If numbers.size is 0, then
we've got an empty list.

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We could just say cout and we could
say unable to determine the smallest,

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list is empty or something like that,
any kind of message that makes sense.

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And we'll put an endline here.

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Now we're in the l section.

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At this point, obviously, there
is some data in that vector so

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what we can do is, we can again
iterate through the entire vector

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and look for the smallest number.

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Well, how do we do that.

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It's pretty simple just pick the
first number and assume it's the

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smallest and then iterate through
the vector and if you see anything

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smaller just replace that 1 with
the smallest 1 yeah so it's again

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it's pretty straightforward.

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So let's do that.

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Let's say we've got a smallest,
we'll call it smallest.

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And what we'll do is we'll
initialize that guy to numbers at 0.

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All right, so let's think
about what's going on here.

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What we're doing here is we're
saying that the smallest integer

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is the first integer in the vector
right now, reasonable assumption.

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We know that there's at
least 1 integer in there.

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So we'll do that.

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Then what we'll say is we'll
do again arrange base for loop.

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We'll say auto, we'll call it num
and the collection is numbers.

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So we'll loop through there and
what we'll say is if the number

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we just read is less than the
smallest that we've already seen

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then what do we do then we make the
smallest the number we just read.

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That's it.

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Really straightforward.

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Now we don't do an output statement
but look where we're going to do it

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we're not going to do it here, we're
going to do it here because at this

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point, we're out of that for loop.

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So we're going to say cout
the smallest number is a

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colon and its smallest, and
we'll provide an endline.

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That's it.

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Let's test this out, and we could
put data in here up top,

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it's right up here just put a bunch of data
and we know what the smallest is.

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But I really want to
test it all as we go.

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So I'm going to run it now.

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And let's add 10, and let's add 20.

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And we'll select s.

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We expect 10.

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That's exactly what we got back.

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Let's add another number.

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Let's add negative 20.

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Let's print the numbers, and
we've got 10, 20 negative 20.

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If we want the smallest at
this point I'll press s again,

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we should get negative 20.

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And there it is, negative 20.

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So a whole bunch of other
test cases you could do,

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but this works just fine.

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I'll press q to quit.

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And next is the largest, right.

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The logic is exactly the same as
this except it's going the other way.

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So in this case let me
clean up that white space.

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Right here, we're going to say
else if the selection is equal

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to a capital l or the selection
is equal to a lowercase l.

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Now the user wants to find
the largest number in there.

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I'm going to use
another block statement.

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And the same logic I'm going to
say in largest and I'm going to

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assume that largest is the number
at the zeroth location, right.

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The first number in the vector.

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Oops, let me back up here.

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I forgot to check to make sure that
there's numbers in there, right.

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So if numbers.size, I got a
little bit ahead of myself there.

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If numbers.size is equal to
zero, then in this case, there

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is nothing to check, right.

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So we'll just do the same
thing we did out here.

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We'll say cout unable to
determine largest, list is empty.

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Here's the else part.

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That's the part where
I got ahead of myself.

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Let's assume that the
largest is that first number.

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I know there is a first number,
right because the size was not 0.

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So at this point, we're going to
loop through the collection again.

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So I'm going to say auto
num, just like we did before.

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And we're going to check if the
number we just read is greater

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than the largest number we've seen,
then what we're going to do is

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we're going to say that the largest
number is now known, simple as that.

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And our output statement
now says the largest number

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is, and we'll say largest.

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Okay, so that should take
care of the largest case.

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So let's test this out.

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So let's add some numbers.

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We'll add negative 10.

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We'll add a 1000.

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We'll add 200, and we'll add 300.

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So let's display our list.

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So you've got negative
10, 1000 200 , 300.

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We expect the largest to be
a 1000, so I'll press l, and

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the largest number is a 1000.

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Let's check the smallest again.

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S, the smallest is negative 10.

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Cool.

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So we're almost done, right.

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All we need to handle is the q case.

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I'll quit here.

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And the q case is real simple because
all we want to do is just say goodbye.

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So let's do that here.

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We'll say else if the selection in
this case, right, just like before is

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an uppercase q or if the selection is
equal to a lowercase q, in this case

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what we want is to simply say cout.

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And let's do that in a block as well.

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You really want to do these in
blocks because you never know

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when you're going to come back
to add code, and it's already

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set up to add code this way.

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So we'll say see out goodbye, endline.

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Perfect.

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And while we're here we may as
well handle that other case right.

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That's the catch catch-all
else, right here.

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That's going to be the end
of the else if else ladder.

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What happens here.

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Well, they typed something in that
I have no idea what to do with.

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So at this point what do we
say,

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cout unknown selection.

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Please try again.

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That's as simple as that.

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Okay, seems like a lot of
code and it sort of is.

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When we talk about functions very
soon, we're going to make this

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so much easier because instead of
all of this code in here,

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we're just going to have one line that
calls a function, and one line that

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calls a function here and so forth.

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So we'll get to there soon. But let's finish this up now.

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Building and running.

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Let's press q, we expect
a goodbye message now.

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There it is, goodbye. And we quit.

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And let's type in something that
we don't know what to do with an o,

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unknown selection, please try again.

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A 1, unknown selection,
please try again.

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And that's it.

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That's the solution.

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Of course, you could have done this
with a switch statement here in which

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case it might have looked something
like switch off of selection.

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And then here you would have case
no Q case lowercase q you know

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cout goodbye, and then break here.

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And then you'd have all
your other cases, just

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like you saw in the videos.

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But in this case, I think the
switch statement is is okay

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if we're using functions.

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We could just call functions here.

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But otherwise, the switch
statement 10ds to get even longer.

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And I don't know how many lines
of code we wrote for this example.

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Let's see, we started on about
line 65 because the rest was just

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the you know the stuff at the top.

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So we went from about
65 down to about 140.

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So I don't know about
70 lines of code or so.

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It would be obviously much
less when we do functions.

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And with the switch statement, I
think it would be even longer the

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amount of code we'd written just
because of the syntax of the switch.

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But the if else if
ladder works great here.

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Obviously, if i decide to add
another option here, it's just

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a matter of coming in here.

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And you could do it anywhere
really because it's all sort

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of mutually exclusive, right.

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And you can just do it right here.

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Else if at it between this one and
this one and there you've got your

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other option just as easy as that.

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Okay, so I hope you
enjoyed this challenge.

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This was a challenging challenge
because it's using everything

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that we've learned so far.

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But there's a lot of logic
going on here, right.

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It's a simple program but there's
a lot of logic going on here.

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You've got you know a do while loop.

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You've got for loops looping through
the iteration, and thank goodness that

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c++11 gave us this range-based for
loop because that makes it so simple.

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We don't have to do things like
for int I equals 0, i less than

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you know the size of the vector
and then increment I along the way.

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We don't have to deal with
any of that, so it makes it

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really really straightforward.

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Okay, so I hope you
enjoyed this section.