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 PyPy3 (7.3.0) (judge verdict: Accepted)

# coding=utf-8
from math import floor, ceil, sqrt, factorial, log, gcd
from itertools import accumulate, permutations, combinations, product, combinations_with_replacement
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from heapq import heappop, heappush, heappushpop, heapify
import copy
import sys
INF = float('inf')
mod = 10**9+7
sys.setrecursionlimit(10 ** 6)


def lcm(a, b): return a * b / gcd(a, b)

# 1 2 3
# a, b, c = LI()


def LI(): return list(map(int, sys.stdin.buffer.readline().split()))

# a = I()


def I(): return int(sys.stdin.buffer.readline())

# abc def
# a, b = LS()


def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()

# a = S()


def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')

# 2
# 1
# 2
# [1, 2]


def IR(n): return [I() for i in range(n)]

# 2
# 1 2 3
# 4 5 6
# [[1,2,3], [4,5,6]]


def LIR(n): return [LI() for i in range(n)]

# 2
# abc
# def
# [abc, def]


def SR(n): return [S() for i in range(n)]

# 2
# abc def
# ghi jkl
# [[abc,def], [ghi,jkl]]


def LSR(n): return [LS() for i in range(n)]

# 2
# abcd
# efgh
# [[a,b,c,d], [e,f,g,h]]


def SRL(n): return [list(S()) for i in range(n)]


n = I()
a = LI()

mx = max(a)
table = [0] * (mx + 1) # num of factors
for x in a:
  table[x] += 1
g = 0
# check 
for x in a:
  g = gcd(g, x)
for i in range(2, mx + 1):
  c = 0
#　因数iの倍数がaにいくつ含まれるか(c)
  for j in range(i, mx + 1, i):
    c += table[j]
# cが2以上=iの倍数となるaが複数存在する
  if c > 1:
    if g == 1:
      print("setwise coprime")
    else:
      print("not coprime")
    exit(0)
print("pairwise coprime")