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;

const int MAXN=1000000;
int A[MAXN+5];
int G[MAXN+5];

int gcd(int a,int b)
{
	return b == 0 ? a : gcd(b,a%b) ;
}

int main()
{
	int n,x;
	int ans=0;
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&x);
		++A[x];
		ans=gcd(ans,x);
	}	
	int ok=1;
	for(int i=2;i<=MAXN;++i)
	{
		int cnt=0;
		for(int j=i;j<=MAXN;j+=i)
		{
			cnt+=A[j];
		}
		if(cnt>1) ok=0;
	}
	if(ok) printf("pairwise coprime");
	else if(ans==1) printf("setwise coprime");
	else printf("not coprime");
	return 0;
} 