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`.

Examples

Input

3
3 4 5


Output

pairwise coprime


Input

3
6 10 15


Output

setwise coprime


Input

3
6 10 16


Output

not coprime
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 1e6 + 100;
typedef long long ll;

//prime次数 
int num[N], n;
inline void divide(int x) {
	for(int i = 2; i <= x/i; i++) {
		if(x % i == 0) {
			num[i]++;
			while(x % i == 0) x /= i;
		}
	}
	if(x > 1) num[x]++;
}
int main() {
	
	cin >> n;
	
	int MaxNum = 0;
	for(int i = 1; i <= n; i++) {
		int x;
		scanf("%d", &x);
		divide(x); 
		
		MaxNum = max(MaxNum, x);
	}
	
	int f1 = 0;
	for(int i = 2; i <= MaxNum; i++) {
		if(num[i] >= n) {
			puts("not coprime");
			return 0;
		}
		if(num[i] > 1) {
			f1 = 1;
			//return 0; 
		}
	}
	if(f1) puts("setwise coprime");
	else puts("pairwise coprime");
	return 0;
}