'in c how preprocessor(#) get evaluated ? not understanding the output of this program [duplicate]

what would be the output of this program ?

#include <stdio.h>
#define s(x)x*x
int main()
{ printf("%d",s(5+1));

      return 0;
}

the output of this code is 11 how?

c logic

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Solution	Source