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: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<cstdio>
#include<cstring>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(n) rep(i,n)
#define all(n) n.begin(),n.end()
#define foreach(it,c) for(typeof(c.begin()) it=c.begin();it!=c.end();++it)

const int MAXM = 170;
int main()
{
    int num; cin >> num;
    
    while(num--)
    {
        int p, q; cin >> p >> q;
        int count = 0, seki = 0;
        for(int m = 0; m < MAXM; m++)for(int n = 0; n < MAXM; n++)
        {
            if((m == 1 && n == 0) ||( m == 0 && n == 1)) continue;
            //cout << m << " " << n << endl;
            seki = m * m + n * n;
            if(seki == 0) continue;
            if(seki >= 20000 ) continue;
            if((((m * p + n * q) % seki) == 0) && (((m * q - n * p) % seki) == 0))
            {
                //if((m == abs(p) && (n == abs(q))) || (m == abs(q) && (n == abs(p)) ) )
                if((m == p && n == q) ||
                        (m == -q && n == p) ||
                        (m == -p && n == -q) ||
                        (m == q && n == -p))
                {
                    continue;
                
                    //cout << m << " " << n << endl;    
                }
                count ++;
            }
        }
        if(count>0) cout << 'C' << endl;
        else cout << 'P' << endl;
    }
    return 0;
}