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 <cstdio>
int gcd(int a,int b){
	if(b==0){
		return a;
	}
	return gcd(b,a%b);
}
const int Maxn=1000000;
int num[Maxn+5];
int n;
int a[Maxn+5];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		num[a[i]]++;
	}
	bool flag=1;
	for(int i=2;i<=Maxn;i++){
		int sum=0;
		for(int j=i;j<=Maxn;j+=i){
			sum+=num[j];
		}
		if(sum>1){
			flag=0;
			break;
		}
	}
	if(flag){
		puts("pairwise coprime");
		return 0;
	}
	int sum=a[1];
	for(int i=2;i<=n;i++){
		sum=gcd(sum,a[i]);
	}
	if(sum==1){
		puts("setwise coprime");
	}
	else{
		puts("not coprime");
	}
	return 0;
}
