Consider the following program:int f(int *p, int n) { if (n <= 1) return 0; else return max(f(p+1,n-1),p[0]-p[1]); } int main() { int a[] = {3,5,2,6,4}; printf("%d", f(a,5)); }Note: $\max(x, y)$ returns the maximum of $x$ and $y$. The value printed by this program is _______.

We can trace the recursive calls of f(p, n) for the array $a = \{3, 5, 2, 6, 4\}$:

1. f(a, 5): evaluates $\max(f(a+1, 4), a[0] - a[1]) = \max(f(a+1, 4), 3 - 5) = \max(f(a+1, 4), -2)$

2. f(a+1, 4): evaluates $\max(f(a+2, 3), a[1] - a[2]) = \max(f(a+2, 3), 5 - 2) = \max(f(a+2, 3), 3)$

3. f(a+2, 3): evaluates $\max(f(a+3, 2), a[2] - a[3]) = \max(f(a+3, 2), 2 - 6) = \max(f(a+3, 2), -4)$

4. f(a+3, 2): evaluates $\max(f(a+4, 1), a[3] - a[4]) = \max(f(a+4, 1), 6 - 4) = \max(f(a+4, 1), 2)$

5. f(a+4, 1): since $n = 1 \le 1$, the base case triggers and returns $0$.

Unwinding the recursion:

• f(a+3, 2)$= \max(0, 2) = 2$

• f(a+2, 3)$= \max(2, -4) = 2$

• f(a+1, 4)$= \max(2, 3) = 3$

• f(a, 5)$= \max(3, -2) = 3$

Hence, the value printed is 3.