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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep3(i, s, n) for (int i = (s); i <= (int)(n); i++)
int b[1000010];
int d[1000010];

int main() {
	int n,a;
	int p,ans;

	cin >> n ;

	ans=0;
	rep(i,1000010) b[i]=0;
	for(int i = 1000005; i > 1; i--) {
		for(int j = i; j<1000005; j+=i) d[j]=i;
	}
	rep(k,n) {
		cin >> a ;
		p=0;
		while (a>1) {
			if(d[a]>p) {
				p=d[a];
				b[p]++;
				if(b[p]>ans) ans=b[p];
			}
			a/=p;
		}
	}
	if (ans==n) {
		cout << "not coprime" << endl;
		return 0;
	}else if (ans>1){
		cout << "setwise coprime" << endl;
		return 0;
	}else{
		cout << "pairwise coprime" << endl;
		return 0;
	}
}