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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 1001001001;
const ll mod = 1000000007;
const int MAX = 1000100;
/**
 * caution : set MAX to the proper value
 * caution : don't forget to use Sieve
 **/
bool Is_prime[MAX];
int MinFactor[MAX];
// Eratosthenes's Sieve
// O(NloglogN)
void Sieve(int n = MAX){
    for (int i = 0; i < n; ++i) Is_prime[i] = true, MinFactor[i] = -1;
    Is_prime[0] = false; Is_prime[1] = false; 
    MinFactor[0] = 0; MinFactor[1] = 1;
    for (int i = 2; i * i < n; ++i) {
        if (Is_prime[i]) {
            MinFactor[i] = i;
            for (int j = i * i; j < n; j += i) {
                Is_prime[j] = false;
                if (MinFactor[j] == -1) MinFactor[j] = i;
            }
        }
    }
}
// Factorization O(logN)
vector<pair<int, int>> prime_factor(int n){
    vector<pair<int,int> > res;
    while (n != 1) {
        int prime = MinFactor[n];
        int exp = 0;
        while (MinFactor[n] == prime) {
            ++exp;
            n /= prime;
        }
        res.push_back(make_pair(prime, exp));
    }
    return res;
}

int gcd(int a, int b){return (b == 0 ? a : gcd(b, a % b));}

int main(){
    int n; cin >> n;
    Sieve();
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    int full = a[0];
    for(int i = 1; i < n; i++){
        full = gcd(a[i], full);
    }
    bool one = true;
    vector<int> d(1000001, 0);
    for(int i = 0; i < n; i++){
        auto pf = prime_factor(a[i]);
        for(auto p : pf){
            d[p.first]++;
        }
    }
    if(*max_element(d.begin(), d.end()) >= 2) one = false;
    if(one) cout << "pairwise coprime" << endl;
    else if(full == 1) cout << "setwise coprime" << endl;
    else cout << "not coprime" << endl;
}