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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
    unsigned long n;
    int loop;
    unsigned long long ans=1;
    // 整数の入力
    scanf("%d", &loop);
    while (loop >= 1)
    {
        scanf("%lu", &n);
        ans = ans * n;
        if (n == 0)
        {
            break;
        }

        loop--;
    }
    // 出力
    if (ans > 1000000000000000000)
    {
        printf("-1");
    }
    else
    {
        printf("%llu",ans);
    }
    return 0;
}