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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define P pair<ll,ll>

const int N=1000010;

ll gcd(ll m, ll n) {//最大公約数
    if (n == 0)return m;
    return gcd(n, m % n);
}//gcd


int main(){
  int n;
  cin>>n;
  vector<int> a(n);
  vector<int> c(N,0);
  rep(i,n){
    cin>>a[i];
    c[a[i]]++;
  }
  bool pairwise=1;
  for(int i=2; i <= N; i++){
    int cnt=0;
    for(int j=i; j <=N; j+=i){
      cnt+=c[j];
    }
    if(cnt>1) pairwise=0;
  }
  if(pairwise){
    cout<<"pairwise coprime"<<endl;
    return 0;
  }
  int p=0;
  rep(i,n){
    p=gcd(p,a[i]);
  }
  if(p==1) cout<<"setwise coprime"<<endl;
  else cout<<"not coprime"<<endl;
}