Trace the output of the following program#include<stdio.h>void main( ){  int K[2][4] = {-40, 36, -55, -88, 65, 74, -99, -24};  printf("%u, %d, %u, %d \n",*K-1,K[1][3]-4, K-1, **K+42); }a.Address 1, -28, Address 2, -2b.Address 1, -28, Address 2, 2c.Address 1, 2, Address 2, -28d.Address 1, -2, Address 2, 28

To trace the output of the given program, let's analyze the code step by step:

1. The program includes the standard input/output library, stdio.h.

2. The main() function is defined.

3. Inside the main() function, a 2-dimensional array K of size 2x4 is declared and initialized with the values {-40, 36, -55, -88, 65, 74, -99, -24}.

4. The printf() function is called to print the output. It contains four placeholders: %u, %d, %u, and %d.

5. The first placeholder, %u, is used to print the value at the address *K-1.

6. The second placeholder, %d, is used to print the value at the address K[1][3]-4.

7. The third placeholder, %u, is used to print the value at the address K-1.

8. The fourth placeholder, %d, is used to print the value at the address **K+42.

Based on the given initialization of the array K, let's calculate the values for each placeholder:

1. The value at the address *K-1 is the value at the address of the first element of the array K minus 1. Since the first element is -40, the value at this address is -41.

2. The value at the address K[1][3]-4 is the value at the address of the element in the second row and fourth column of the array `