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: Time Limit Exceeded)

#include<stdio.h>

int main(){
	unsigned long long int batas,sum;
	sum=1;
	batas= 10000000000000000000;
	long int i,n;
	scanf("%d", &n);
	for(i=1;i<=n;i++){
		unsigned long long int angka;
		scanf("%d", &angka);
		if(angka==0){
			sum=0;
		}
		else{
			sum*=angka;
			//printf("jumlah sum ke %d adalah %lld\n", i, sum);
		}
	}
	if(sum>batas){
		printf("-1\n");
		//printf("Batas adalah %lld", batas);
	}
	else{
		printf("%lld\n", sum);
		//printf("Batas adalah %lld", batas);
	}
	return 0;
}