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: Accepted)

import numpy as np
from numba import njit
def main():
    N = int(input())
    nlist = np.array(list(map(int, input().split())))

    def gdc(x, y):
        b = max(x, y)
        s = min(x, y)
        r = b % s
        if r == 0:
            return s
        else:
            return gdc(s, r)

    scp = False
    pcp = False

    sgdc = nlist[0]
    for n in nlist:
        sgdc = gdc(sgdc, n)
        if sgdc == 1:
            scp = True

    if not scp:
        print('not coprime')
        exit()

    pcheckl = [0] * (10**6+1)
    for n in nlist:
        pcheckl[n] = 1

    pcp = all([sum(pcheckl[i::i]) <= 1 for i in range(2, 10**6+1)])

    if pcp:
        print('pairwise coprime')
    elif scp:
        print('setwise coprime')

if __name__ == '__main__':
    main()