Given an array $A[p..r],$ the following recursive algorithm Min$(A,p,r)$ will return the smallest element in $A[p.r].$ The algorithm recursively divide the input array into two subarrays and compare the minimums of the two subarrays to find the smallest element of the original array $A[p..r].$

Min$(A,p,r)$

if $p==r$

return $A[p]$

else

$q=\frac{p+r}{2}$

$a=$Min$(A,p,q)$

$b=$Min$(A,q+1,r)$

if $A[1.\mathrm{.}6]=($:$6,9,5,8,2,3$:$)(A,1,6)Min(A,1,3)Min(A,4,5)Min(A,2,2)Min(A,5,6)a$

return $a$

else

return $b$

Given $an$ array $A[1.\mathrm{.}6]=($:$6,9,5,8,2,3$:$),$ consider calling Min $(A,1,6).$ Which $of$ the following will NOT $be$ a recursive call $to$ Min during the execution $of$ the procedure?

Min$(A,1,3)$

Min$(A,4,5)$

Min$(A,2,2)$

Min$(A,5,6)$