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>
#include <math.h>

int main(){
    int data[100000];
    int N,num,i,n,sum;
    //char s[20];
    num = 0;
    sum = 1;

    scanf("%d",&N);
    for(i=0;i<N;i++){
        scanf("%d", &data[i]);
    }

    for(num=0;num<N;num++){
        sum = sum * data[num];
    }

    //sprintf(s,"%d",sum);

   // for(n =0;s[n] != '\0' ; n++)

    if(sum <= pow(10,18)){
        printf("%d",sum);
    }else{
        printf("%d",-1);
    }

    return 0;

}