Consider the following C program.#include <stdio.h> int main() { int a[4][5]={{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18 , 19, 20}}; printf("%d ", *(*(a+**a+2)+3)); return(0); }The output of the program is _________.

Let us evaluate the expression *(*(a + **a + 2) + 3) step by step:

1. Evaluate `a`**:

   • a is a 2D array of type int[4][5].

   • In an expression, a decays to a pointer to its first row, int (*)[5].

   • *a gives the first row a[0], which decays to a pointer to a[0][0] (type int *).

   • **a dereferences *a, yielding the value at a[0][0], which is $1$.

2. Evaluate `a + a + 2`**:

   • Substitute **a = 1:

     $$\text{offset} = 1 + 2 = 3$$

   • a + 3 points to the 4th row, a[3] (type int (*)[5]).

3. **Evaluate *(a + 3) + 3**:

   • *(a + 3) is equivalent to a[3], which decays to &a[3][0] (type int *).

   • Adding $3$ shifts the pointer to &a[3][3].

4. **Evaluate *(*(a + 3) + 3)**:

   • Dereferencing gives a[3][3].

   • Looking at row index 3: {16, 17, 18, 19, 20}:

     • a[3][0] = 16

     • a[3][1] = 17

     • a[3][2] = 18

     • a[3][3] = 19

Therefore, the program prints 19.