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
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ALL(v) v.begin(), v.end()
typedef long long ll;

#include <bits/stdc++.h>
using namespace std;

int A[1000100];

int main(){
  int n;
  cin>>n;
  
  rep(j,n){
    int x;
    cin>>x;
    set<int> s;
    for(int i=2;i*i<=x;i++){
      while(x%i==0){
        s.insert(i);
        x/=i;
      }
    }
    if(x!=1) s.insert(x);
    
    for(auto y:s){
      A[y]++;
    }
  }
  
  int ma=0;
  for(int i=2;i<=1000000;i++){
    ma=max(ma,A[i]);
  }
  
  if(ma==n) cout<<"not coprime"<<endl;
  else if(ma>1) cout<<"setwise coprime"<<endl;
  else cout<<"pairwise coprime"<<endl;

  return 0;
}
