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<algorithm>
#include<iostream>
#include<vector>
#include<set>

using namespace std;

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

int main()
{
	int n;
	cin>>n;
	vector<int> a(n);
	set<int> s;
	for(int i=0;i<n;i++)  scanf("%d",&a[i]);
	int gc=gcd(a[0],a[1]);
	for(int i=2;i<n;i++)	gc=gcd(gc,a[i]);
	if(gc!=1)
	{
		cout<<"not coprime"<<endl;
		return 0;
	}
	for(int i=0;i<n;i++)
	{
		if(a[i]==1)	continue;
		for(int j=1;j*j<=a[i];j++)
		{
			if(a[i]%j==0)
			{
				if((j!=1 && s.count(j)) || s.count(a[i]/j))
				{
					cout<<"setwise coprime"<<endl;
					return 0;
				}
				s.insert(j);
				s.insert(a[i]/j);
			}
		}
	}
	cout<<"pairwise coprime"<<endl;
	return 0;
}