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>
using namespace std;
typedef long long ll;
const long long INF = 1LL << 60; //intじゃ扱えないことに注意！
#define mod 1000000007
#define rep(i, n) for(int i = 0; i < (int)(n); i++) //範囲外参照とループの初期化に注意！
const double pi = 3.14159265359;


int main() {
  ll N;
  cin >> N;
  ll ans = 1;
  vector<ll> A(N);
  for(int i = 0; i < N; i++){
      cin >> A[i];
  }
  for(int i = 0; i < N; i++){
      ans *= A[i];
      if(ans > 1000000000000000000){
          break;
      }
  }
  
  if(ans > 1000000000000000000){
      cout << -1 << endl;
  }
  else{
      cout << ans << endl;
  }
}