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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>

signed main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);

    int n;
    std::cin >> n;

    std::vector<int> a(n);
    for (int &i : a) std::cin >> i;

    int gcd = 0;
    for (int i : a) gcd = std::gcd(gcd, i);

    if (gcd != 1) {
        std::cout << "not coprime";
        return 0;
    }

    int mulOf2 = 0;
    for (int i : a) mulOf2 += (i % 2 == 0);

    if (mulOf2 > 1) {
        std::cout << "setwise coprime";
        return 0;
    }

    for (int i = 3; i <= (int) 1e6; i += 2) {
        int mulOfi = 0;
        for (int x : a) mulOfi += (x % i == 0);
        if (mulOfi > 1) {
            std::cout << "setwise coprime";
            return 0;
        }
    }

    std::cout << "pairwise coprime";

    return 0;
}