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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
int gcd(int x,int y){
  if(x%y == 0) return y;
  else return gcd(y,x%y);
}
int A[1000000];
int main(){
  int N,i,j,k,a;scanf("%d",&N);
  int B[N];
  for(i=0;i<N;i++){
    scanf("%d",&a);
    B[i] = a;
    A[a]++;
  }
  bool ok = true;int t;
  for(i=2;i<=1000000;i++){
    t = 0;
    for(j=i;j<=1000000;j+=i){
      t += A[j];
    }
    ok &= t<=1;
  }
  if(ok){
    printf("pairwise coprime\n");return 0;
  }
  int ans = B[0];
  for(i=1;i<N;i++) ans = gcd(ans,B[i]);
  printf(ans == 1 ? "setwise coprime\n":"not coprime\n");
}