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>
using namespace std;

int main() {
    long long a[100010];
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    int zero = 0;
    for (int i = 0; i < n; i++) {
        if (a[i] == 0) zero++;
    }
    if (zero > 0) {
        cout << "0" << endl;
        return 0;
    }

    long long prod = 1;
    for (int i = 0; i < n; i++) {
        if (a[i] <= 1000000000000000000 / prod) {
            prod *= a[i];
        } else {
            cout << -1 << endl;
            return 0;
        }
    }
    cout << prod << endl;
}