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
from math import gcd
from functools import reduce
def osa_k(n=10**6):
  p=list(range(n+1))
  for i in range(2,int(n**.5)):
    if p[i]<i:continue
    for j in range(i*i,n+1,i):
      if p[j]==j:p[j]=i
  return p
def fact_o(n):
  a=[]
  f=0
  while n>1:
    if f!=p[n]:ans[p[n]]+=1
    f=p[n]
    n//=p[n]
  return 0
p=osa_k()
n=int(input())
a=list(map(int,input().split()))
ans=[0]*(10**6+1)
for i in a:
  fact_o(i)
if max(ans)<=1:print('pairwise coprime')
elif reduce(gcd,a)==1:print('setwise coprime')
else:print('not coprime')