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;

int n, x[1000005], cnt[1000005];
bool np[1000005];

int gcd(int a, int b) {
	while (a) {
		b %= a;
		swap(a, b);
	}
	return b;
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", x + i);
		cnt[x[i]]++;
	}
	int g = x[0];
	for (int i = 1; i < n; i++)
		g = gcd(g, x[i]);
	bool ok = 1;
	for (int i = 2; i < 1000005; i++)
		if (!np[i]) {
			int tot = 0;
			for (int j = i; j < 1000005; j += i) {
				tot += cnt[j];
				np[j] = 1;
			}
			if (tot > 1)
				ok = 0;
		}
	if (ok)
		puts("pairwise coprime");
	else if (g == 1)
		puts("setwise coprime");
	else
		puts("not coprime");
}