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>
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define int long long
using namespace std;
 
const int N = 2e6 + 5;
const int mod = 1e9 + 7;

int a[N];
int occ[N];

int32_t main() {
	fast;
	int n;
	cin>>n;
	for(int i=0; i<n; i++) cin>>a[i];
	bool ok = true;
	for(int i=0; i<n; i++){
		int x = a[i];
		for(int i=2; i*i <=x; i++){
			if(x%i == 0){
				occ[i]++;
				if(occ[i]>1) ok = false;
				while(x%i == 0 ) x/= i;
			}
		}
		if(x!=1) occ[x]++;
		if(occ[x] > 1) ok = false;
	}
	if(ok){
		cout<<"pairwise coprime" <<endl;
		return 0;
	}
	for(int i=1; i<= 1000000; i++){
		if(occ[i] == n){
			cout<<"not coprime" <<endl;
			return 0;
		}
	}
	cout<<"setwise coprime"<<endl;
}



