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 <stdio.h>

int main(void){
    long long int n;
    scanf("%lld",&n);
    long long int a;
    scanf("%llu",&a);
    long long int ans = a;
    long long int ai[n-1];
    long long int i=0;
    if(ans == 0) {
        printf("0\n");
        return 0;
    }
    for(i = 0;i<n-1;i++){
        scanf("%llu",&ai[i]);
        if(ai[i] == 0) {
            printf("0\n");
            return 0;
        }
        if(ai[i] > 1000000000000000000){
            printf("-1\n");
            return 0;           
        }
    }

    for(i = 0;i<n-1;i++){
        if(ans>(ans*ai[i])){
            printf("-1\n");
            return 0;
        }
        ans *= ai[i];
        if(ans>1000000000000000000) {
            printf("-1\n");
            return 0;
        }
    }

    printf("%lld\n",ans);
    return 0;
}