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 N=1e6;
int A[N+1];
int GCD(int a,int b){
	return b==0?a:GCD(b,a%b);
}
int main(){
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
#endif
	int n,tmp,gcd=0;
	cin>>n; 
	for(int i=0;i<n;++i){
		cin>>tmp;
		++A[tmp];
		gcd=GCD(tmp,gcd);
	}
	
	//pairwise
	for(int i=2;i<=N;++i){
		int flag=0;
		for(int j=i;j<=N;j+=i){
			if(j<=N && A[j]>=1) flag+=A[j];
			if(flag>=2) goto setwise;
		}
	}
	cout<<"pairwise coprime"<<endl;
	return 0;
	
	setwise:
	if(gcd==1) cout<<"setwise coprime"<<endl;
	else cout<<"not coprime"<<endl;
	return 0;
}
