Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

3
3 4 5

Sample Output 1

pairwise coprime

GCD(3,4)=GCD(3,5)=GCD(4,5)=1, so they are pairwise coprime.

Sample Input 2

3
6 10 15

Sample Output 2

setwise coprime

Since GCD(6,10)=2, they are not pairwise coprime. However, since GCD(6,10,15)=1, they are setwise coprime.

Sample Input 3

3
6 10 16

Sample Output 3

not coprime

GCD(6,10,16)=2, so they are neither pairwise coprime nor setwise coprime.

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

from sys import stdin,stdout
from math import gcd
n=int(stdin.readline())
a=list(map(int,stdin.readline().split()))
s=set(a)
if len(s)!=n:
    print('not coprime')
    exit(0)
p=2;d={}
N=max(a)+2
primes=[1]*N;not_pairwise=0
while p*p<=N:
    if primes[p]:
        for mul in range(p,N,p):
            primes[mul]=0
            if mul in s:
                d[p]=d.get(p,0)+1
                if d[p]>=2:not_pairwise=1;break;
        if not_pairwise:break
    p+=1
g=a[0]
for v in a:
    g=gcd(g,v)
if not_pairwise==0:print('pairwise coprime')
elif g==1:print('setwise coprime')
else:print('not coprime')