Problem:
Problem A: Ginkgo Numbers

We will define Ginkgo numbers and multiplication on Ginkgo numbers.

A Ginkgo number is a pair <m, n> where m and n are integers. For example, <1, 1>, <-2, 1> and <-3,-1> are Ginkgo numbers.

The multiplication on Ginkgo numbers is defined by <m, n> · <x, y> = <mx − ny, my + nx>. For example, <1, 1> · <-2, 1> = <-3,-1>.

A Ginkgo number <m, n> is called a divisor of a Ginkgo number <p, q> if there exists a Ginkgo number <x, y> such that <m, n> · <x, y> = <p, q>.

For any Ginkgo number <m, n>, Ginkgo numbers <1, 0>, <0, 1>, <-1, 0>, <0,-1>, <m, n>, <-n,m>, <-m,-n> and <n,-m> are divisors of <m, n>. If m2+n2 > 1, these Ginkgo numbers are distinct. In other words, any Ginkgo number such that m2 + n2 > 1 has at least eight divisors.

A Ginkgo number <m, n> is called a prime if m2+n2 > 1 and it has exactly eight divisors. Your mission is to check whether a given Ginkgo number is a prime or not.

The following two facts might be useful to check whether a Ginkgo number is a divisor of another Ginkgo number.

Suppose m2 + n2 > 0. Then, <m, n> is a divisor of <p, q> if and only if the integer m2 + n2 is a common divisor of mp + nq and mq − np.

If <m, n> · <x, y> = <p, q>, then (m2 + n2)(x2 + y2) = p2 + q2.

Input

The first line of the input contains a single integer, which is the number of datasets.

The rest of the input is a sequence of datasets. Each dataset is a line containing two integers m and n, separated by a space. They designate the Ginkgo number <m, n>. You can assume 1 < m2 + n2 < 20000.

Output

For each dataset, output a character 'P' in a line if the Ginkgo number is a prime. Output a character 'C' in a line otherwise.

Sample Input

8
10 0
0 2
-3 0
4 2
0 -13
-4 1
-2 -1
3 -1

Output for the Sample Input

C
C
P
C
C
P
P
C

Solution in C++ (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdlib>
using namespace std;
 
bool D(int p,int q,int m,int n)
{
  if((m*p+n*q)%(m*m+n*n)==0 && (m*q-n*p)%(m*m+n*n)==0)return true;
  return false;
}
 
bool judge(int p,int q)
{
  int c=0;
  for(int i=1;i*i<=p*p+q*q;i++){
    for(int j=i;j*j<=p*p+q*q;j++){
      if(i!=abs(q) && j!=abs(p)){
	if(D(p,q,i,j))c++;
	if(D(p,q,i,-j))c++;
	if(D(p,q,-i,j))c++;
	if(D(p,q,-i,-j))c++;
      }
    }
  }
  if(c==0)return true;
  return false;
}
 
int main()
{
  int n,m,N;
  cin>>N;
  for(int i=0;i<N;i++){
    cin>>m>>n;
    if(judge(m,n))cout<<"P"<<endl;
    else cout<<"C"<<endl;
  }
  return 0;
}