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()
{
	long long sum = 1,ans;
	int flagO = 1,flag = 1;
	long long tc;
	scanf("%lld",&tc);
	long long input[tc+1];
	for(int i = 0;i<tc;i++)
	{
		scanf("%lld",&input[i]);
		if(input[i] == 0) 
		{
			flagO = 0;	
		}	
	}
	for(int i = 0;i<tc;i++)
	{
		if(sum *input[i] > 1000000000000000000 || sum * input[i] < 0 || input[i] > 1000000000000000000) 
		{
			flag = 0;
			break;
		}
		sum *= input[i];
	}
	if(flagO == 0) printf("0\n");
	else if(flag == 0) printf("-1\n");
	else printf("%lld",sum); 
 
}