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: Wrong Answer)

#include<stdio.h>
int main()
{
    int n;
   
    if(scanf("%d", &n)<1){
       return 0;
    }
    
    long long u,a[n],r;
    for(long long i=0;i<n;i++){
        if(scanf("%llu", &u)<1){
            return 0;
        }
        a[i]=u;
    }
    
    for(long long j=0;j<n;j++){
        if(j==0){
            r=a[0];
        }else{
            if(a[j]==0){
                printf("0");
                return 0;
            }else{
                r=r*a[j];
            }
        }
    }
    
    if(r > 1000000000000000000){
        r = -1;
    }
    printf("%llu",r);
   
    return 0;
}