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 REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;
const int MAXI = 1e6 + 10;

template<class T>
inline T gcd(T a, T b) {
    if (b == 0) return a;
    return  gcd(b, a % b);
}

int main() {
    cin.tie(0)->sync_with_stdio(false);

	int n; cin >> n;
	vector<int> cnt(MAXI, 0);
	int d = 0;
	REP(i, n) {
		int a; cin >> a;
		cnt[a]++;
		d = gcd(d, a);
	}
	if (d > 1) {
		cout << "not coprime" << '\n';
		return 0;
	}

	for (int i = 2; i < MAXI; i++) {
		int sum = 0;
		for (int j = i; j < MAXI; j += i)
			sum += cnt[j];
		if (sum > 1) {
			cout << "setwise coprime" << '\n';
			return 0;
		}
	}

	cout << "pairwise coprime" << '\n';
    return 0;
}
