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++11 (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
using namespace std;

int main() {
  int X; cin >> X;
  for(int i = 0; i < X; ++i) {
    int N, M; cin >> N >> M;
    if(N*N + M*M <= 1) {
      cout << "C" << endl;
      continue;
    }
    vector<pair<int,int>> ans;
    for(int a = -200; a <= 200; ++a) {
      for(int b = -200; b <= 200; ++b) {
        int S = a*N + b*M;
        int T = a*M - b*N;
        int U = a*a + b*b;
        if(U == 0) continue;
        if(S % U == 0 && T % U == 0) {
          ans.emplace_back(a, b);
        }
      }
    }
    sort(ans.begin(), ans.end());
    ans.erase(unique(ans.begin(), ans.end()), ans.end());
    if(ans.size() == 8) {
      cout << "P" << endl;
    }
    else {
      cout << "C" << endl;
    }
  }
}