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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int(i) = 0; (i) < (n); (i)++)
#define FOR(i, m, n) for(int(i) = (m); (i) < (n); (i)++)
#define All(v) (v).begin(), (v).end()
#define pb push_back
#define MP(a, b) make_pair((a), (b))
template <class T> vector<T> make_vec(size_t a, T val) {
    return vector<T>(a, val);
}
template <class... Ts> auto make_vec(size_t a, Ts... ts) {
    return vector<decltype(make_vec(ts...))>(a, make_vec(ts...));
}
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using Graph = vector<vector<int>>;
template <typename T> struct edge {
    int to;
    T cost;
    edge(int t, T c) : to(t), cost(c) {}
};
template <typename T> using WGraph = vector<vector<edge<T>>>;
const int INF = 1 << 30;
const ll LINF = 1LL << 60;
const int MOD = 1e9 + 7;

// 素数篩＋素因数分解
// 初期化O(N),素因数分解O(logN)
// たくさん素因数分解するときはこっち
#include <cassert>
struct prime_sieve {
    int N;
    vector<ll> sieve;
    prime_sieve(int n) : N(n + 1) {
        sieve.resize(N);
        iota(All(sieve), 0);
        init();
    }

    void init() {
        for(int i = 2; i * i <= N; i++) {
            if(sieve[i] < i)
                continue;
            for(int j = i * i; j < N; j += i) {
                if(sieve[j] == j)
                    sieve[j] = i;
            }
        }
    }

    map<ll, int> prime_factorize(ll n) {
        assert(n <= N);
        map<ll, int> ret;
        while(n > 1) {
            ret[sieve[n]]++;
            n = n / sieve[n];
        }
        return ret;
    }
};
int main() {
    int N;
    cin >> N;
    vector<ll> A(N);
    rep(i, N) { cin >> A[i]; }
    ll g = A[0];
    rep(i, N) { g = __gcd(g, A[i]); }
    if(g != 1) {
        cout << "not coprime" << endl;
    } else {
        prime_sieve sieve(1000001);
        set<int> se;
        bool ok = 1;
        rep(i, N) {
            auto m = sieve.prime_factorize(A[i]);
            for(auto p : m) {
                if(se.count(p.first))
                    ok = 0;
                se.insert(p.first);
            }
        }
        if(ok) {
            cout << "pairwise coprime" << endl;
        } else {
            cout << "setwise coprime" << endl;
        }
    }
}