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)

/***
*  author  :_shamim
*  created : 07.05.2020
*
***/

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <list>
#include <set>
#include <queue>
#include <iterator>
#include <bitset>
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#define ll             long long
#define _              ios_base::sync_with_stdio(0);cin.tie(0);
#define loop           for(i=0;i<n;i++)
#define pb             push_back
#define tc()           int t;cin>>t;while(t--) 
#define pi             acos(-1)         

using namespace std;

ll i,j,temp;

int main()
{_
    int n;
			long long unsigned int x=1,arr[n];
			cin>>n;
			for(i=0;i<n;i++){
						cin>>arr[i];
			}
			for(i=0;i<n;i++){

						x*= arr[i];
						if(x>1000000000000000000){
									cout<<-1<<endl;
									return 0;
						}
			}
			cout<<x<<endl;


return 0;
}