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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define P pair<ll, ll>
#define Graph vector<vector<ll>>
#define fi first
#define se second
constexpr ll mod = 1000000007;
constexpr ll INF = (1ll << 60);
constexpr double pi = 3.14159265358979323846;
template <typename T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

ll gcd(ll a, ll b) {
    if (b == 0) {
        return a;
    } else {
        return gcd(b, a % b);
    }
}

vector<ll> sum;
void pre_process(vector<ll> &a, vector<ll> &s) {  //累積和のvector
    ll n = (ll)a.size();
    s.assign(n + 1, 0);
    s[0] = a[0];
    for (ll i = 0; i < n; i++) {
        s[i + 1] = gcd(s[i], a[i]);
    }
}

vector<ll> prime_num(5000010, 0);
vector<bool> prime(5000010, true);
bool ans = false;
void isprime(ll N) {
    if (N >= 0) prime[0] = false;
    if (N >= 1) prime[1] = false;
    for (ll i = 2; i * i <= N; i++) {
        if (!prime[i]) {
            continue;
        }

        ll cnt = 0;
        cnt += prime_num[i];

        for (ll j = i * i; j <= N; j += i) {
            prime[j] = false;
            cnt += prime_num[j];
        }

        // if (i == 2) cout << cnt << endl;
        if (cnt >= 2) {
            ans = true;
            break;
        }
    }
}

int main() {
    ll n;
    cin >> n;
    vector<ll> a(n);
    bool ans1 = false;
    ll number = 1;
    rep(i, n) {
        cin >> a[i];

        if (i == 0)
            number = a[i];
        else
            number = gcd(number, a[i]);

        prime_num[a[i]]++;
    }

    isprime(5000000);
    if (number == 1) {  // gcd=1;

        if (ans == true) {
            cout << "setwise coprime\n";
        } else {
            cout << "pairwise coprime\n";
        }
    } else {
        cout << "not coprime\n";
    }

    return 0;
}