Predict the output of the following code snippet.123456789#include <stdio.h>int main() { int x = 5, y = 3; int *ptr1 = &x, *ptr2 = &y; int a = *ptr1 * *ptr2; printf("%d\n", x); return 0;}

The output of the code will be "5".

Here's the step by step explanation:

1. The code starts with the declaration of two integer variables, x and y, with values 5 and 3 respectively.
2. Then, two integer pointers ptr1 and ptr2 are declared and initialized with the addresses of x and y respectively.
3. After that, an integer a is declared and initialized with the product of the values pointed by ptr1 and ptr2. However, this variable a is not used anywhere in the code.
4. The printf function is then called to print the value of x, which is 5.
5. The program ends with a return statement returning 0, indicating successful termination of the program.

So, the output of this code will be "5".