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 <bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(long long i=a; i<b; i+=1)
#define repr(i,a,b) for(long long i=a; i<=b; i+=1)
#define vec vector<ll>
#define map map<char,int>
#define repa(p,A) for(auto p:A)
#define pb push_back
#define sort(a) sort(a.begin(),a.end())
#define reverse(a) reverse(a.begin(),a.end())
const double PI=acos(-1);
using namespace std;

int main( ) {
    ll N;
    cin>>N;
    ll count=1;
    vec A(N);
    rep(i,0,N) cin>>A[i];
    sort(A);
    rep(i,0,N) {
        count*=A[i];
        if(count==0) {
            cout<<"0"<<endl;
            return 0;
        }
        if(count>1000000000000000000) {
            cout<<"-1"<<endl;
            return 0;
        }
        }
        cout<<count<<endl;
        }