Problem:
Score : 500 points

Problem Statement

We have N integers. The i-th number is A_i.

\{A_i\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.

\{A_i\} is said to be setwise coprime when \{A_i\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.

Determine if \{A_i\} is pairwise coprime, setwise coprime, or neither.

Here, GCD(\ldots) denotes greatest common divisor.

Constraints

2 \leq N \leq 10^6

1 \leq A_i\leq 10^6

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

If \{A_i\} is pairwise coprime, print pairwise coprime; if \{A_i\} is setwise coprime, print setwise coprime; if neither, print not coprime.

Sample Input 1

3
3 4 5

Sample Output 1

pairwise coprime

GCD(3,4)=GCD(3,5)=GCD(4,5)=1, so they are pairwise coprime.

Sample Input 2

3
6 10 15

Sample Output 2

setwise coprime

Since GCD(6,10)=2, they are not pairwise coprime. However, since GCD(6,10,15)=1, they are setwise coprime.

Sample Input 3

3
6 10 16

Sample Output 3

not coprime

GCD(6,10,16)=2, so they are neither pairwise coprime nor setwise coprime.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <set>
#include <map>
#define rep(i, n)   for(int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

map<int, int>m;

void prime_fac(int x){
    if(x == 1)  return;
    for(int i = 2; x > 1; i++){
        if(x%i == 0){
            m[i]++;
            while(x%i == 0){
                x /= i;
            }
        }
    }
}

int main(){
    int n;  cin >> n;
    vector<int>a(n);
    rep(i, n)   cin >> a[i];
    rep(i, n)   prime_fac(a[i]);
    int flag = 0;
    for(int i = 2; i <= *max_element(a.begin(), a.end()); i++){
        flag = max(flag, m[i]);
    }
    if(flag == 1){
        puts("pairwise coprime");
    }else if(flag == n){
        puts("not coprime");
    }else{
        puts("setwise coprime");
    }
}