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: Runtime Error)

#include <bits/stdc++.h>
#define D(x) cout << #x << "=" << x <<endl; 
using namespace std; 

//int _; 

int n;
long long ans=1,a[100050];

int main()
{
	//for(scanf("%d",&_);_;_--)
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%lld",a+i);
		if(!a[i])
		{
			puts("0");
			return 0;
		}
	}
	for(int i=0;i<n;i++)
	{
		if(1000000000000000000/ans >= a[i] && 1000000000000000000%ans != 0)
		{
			puts("-1");
			return 0;
		}
		ans *= a[i];
	}
	cout << ans << endl;	 
	return 0;
}