Boring factorials

problem is here

This can be solved using Wilson theorem
   (1). if n>=p  ans would be 0
   (2). then we have to use Wilson's theorem
             (p-1)!   ≡  -1 (mod p)
             1*2*3*.........*(n-1)*(n)*..............*(p-1)  ≡   -1 (mod p)
             n!*(n+1)*...........*(p-1)   ≡   -1 (mod p)
             n!  ≡    -1*[(n+1)*...............(p-2)*(p-1)]^-1 (mod p)


#include<stdio.h>
long long power(long long a,long long b,long long m){
  long long x=1,y=a;
  while(b>0){
  if(b%2!=0){
  x=(x*y)%m;
  }
  y=(y*y)%m;
  b>>=1;
  }
  return x;
}
int main(){
  int t;
  long long n,p,i,result,z,x;
  scanf("%d",&t);
  while(t--){
  result=-1;
  scanf("%lld %lld",&n,&p);
  if(n>=p){
    printf("0\n");
    continue;
  }
  x=1;
  for(i=n+1;i<p;i++){
  x=(x*i)%p;
  }
    z=power(x,p-2,p);
    result=(result*z)%p;
  printf("%lld\n",p+result);
  }
  return 0;

}