(Operator pre: * / %) Why 16?

Question

#include <stdio.h>
#define sqr(x) x*x

int main() {
printf("%d", 16/sqr(4));
return 0;
}

Answer 1

16/sqr(4) = 16/4*4
Try this:
#define sqr(x) (x*x)
or:
printf("%d", 16/(sqr(4)));
this equal 16/(4*4)

Answer 2

16/4*4=16 But, in operator precedence, * first then / then %. Does it should = 1? 16/4*4 => 16/(4*4).