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++)
#define sz(x) int(x.size())
using namespace std;
typedef long long ll;
const long long  mod = 1000000007;


int main(){
  int n;
  cin >> n;

  vector<int> a(1000001,0);
  rep(i,n) {
    int A;
    cin >> A;
    a[A] += 1;
  }

  //pairwise
  bool flag1 = 1;
  //setwise
  bool flag2 = 1;

  for(int i = 2; i <= 1000000; i++){
    int cnt = 0;
    for(int j = i; j <= 1000000; j += i){
      if(a[j] > 0) cnt += a[j];
    }
    if(cnt >= 2) flag1 = 0;
    if(cnt == n) flag2 = 0;
  }
  
  if(flag1) cout << "pairwise coprime" << endl;
  else if (flag2) cout << "setwise coprime" << endl;
  else cout << "not coprime" << endl;
  
 }
