#include using namespace std; int sum; const int num$\_$threads $=4$; void foo$($int val$)\{$ for$($int i $=0$; i $<5$; i$++)\{$ sum $=$ sum $+$ val; $\}\}$ void bar$\_1()\{$ thread t$[$num$\_$threads$]$; for $($int i $=0$; i $<$ num$\_$threads; i$++)\{$ t$[$i$]=$ thread$($foo$,$ i $+1)$; t$[$i$].$join$()$; $\}$ cout $<<$ "Sum is $"<<$ sum $<<$ endl; $\}$ void bar$\_2()\{$ thread t$[$num$\_$threads$]$; for $($int i $=0$; i $<$ num$\_$threads; i$++)\{$ t$[$i$]=$ thread$($foo$,$ i $+1)$; $\}$ cout $<<$ "Sum is $"<<$ sum $<<$ endl; for $($int i $=0$; i $<$ num$\_$threads; i$++)\{$ t$[$i$].$join$()$; $\}\}$ void bar$\_3()\{$ thread t$[$num$\_$threads$]$; for $($int i $=0$; i $<$ num$\_$threads; i$++)\{$ t$[$i$]=$ thread$($foo$,$ i $+1)$; $\}$ for $($int i $=0$; i $<$ num$\_$threads; i$++)\{$ t$[$i$].$join$()$; $\}$ cout $<<$ "Sum is $"<<$ sum $<<$ endl; $\}$ int main$()\{$ sum $=0$; bar$\_1()$; sum $=0$; bar$\_2()$; sum $=0$; bar$\_3()$; $\}$ a$)$ The code on the left outputs three values for sum $($via the cout statements in the bar$\_1,$ bar$\_2,$ and bar$\_3$ functions$).$ sum is a global variable and see that its value is set to $0$ before each call of bar functions. What is the maximum possible value one can see on the screen? $($Hint: The maximum is the same for all outputs$).\_\_\_\_\_\_\_\_\_\_\_\_\_$ b$)(10$ pts$)$ What is the minimum possible value for each output? $($Hint: The minimum values are different for different bar function calls$).$ bar$\_1$: Sum is $($Min$)\_\_\_\_\_\_\_\_$ bar$\_2$: Sum is $($Min$)\_\_\_\_\_\_\_\_$ bar$\_3$: Sum is $($Min$)\_\_\_\_\_\_\_\_$