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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>

using namespace std;
typedef long long ll;

int main(){
    ll n;
    cin>>n;
    vector<ll> a(n);
    for(ll i=0;i<n;i++){
        cin>>a[i];        
    }
    auto z=find(a.begin(),a.end(),0);
    if(z!=a.end()){
        cout<<0<<endl;
    }else{
        ll ans=1;
        for(ll i=0;i<n;i++){
            string s1=to_string(ans);
            string s2=to_string(a[i]);            
            if(s1.length()+s2.length()-2>18){
                ans=-1;
                break;
            }
            ans*=a[i];
        }
        if(ans<=1000000000000000000){
            cout<<ans<<endl;
        }else{
            cout<<-1<<endl;
        }
        
    }
    
    return 0;
}