Problem:
Score : 200 points

Problem Statement

Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.

However, if the result exceeds 10^{18}, print -1 instead.

Constraints

2 \leq N \leq 10^5

0 \leq A_i \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the value A_1 \times ... \times A_N as an integer, or -1 if the value exceeds 10^{18}.

Sample Input 1

2
1000000000 1000000000

Sample Output 1

1000000000000000000

We have 1000000000 \times 1000000000 = 1000000000000000000.

Sample Input 2

3
101 9901 999999000001

Sample Output 2

-1

We have 101 \times 9901 \times 999999000001 = 1000000000000000001, which exceeds 10^{18}, so we should print -1 instead.

Sample Input 3

31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0

Sample Output 3

0

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
int dig(ll n) {
    int i = 0;
    while(true) {
        n/=10;
        if(n == 0) return i;
        i++;
    }
}

int main(){
    ios::sync_with_stdio(false);
    const ll MAX = 1000000000000000000;
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i,n) {
        cin >> a[i];
        if(a[i] == 0) {
            cout << 0 << endl;
            return 0;
        }
    }

    ll ans=1;
    rep(i,n) {
        if(ans*a[i] != MAX && dig(ans)+dig(a[i]) >= 18) {
            cout << -1 << endl;
            return 0;
        }
        ans*=a[i];
        if(ans > MAX) {
            cout << -1 << endl;
            return 0;
        }
    }
    cout << ans << endl;
}