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)

/**
 *  winners never quit
**/

#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
typedef long long Long;
void FastIO(){
  ios::sync_with_stdio(0);
  cin.tie(0);cout.tie(0);
}

const int N = 1e6;

int d(Long n){
  if (n==0)return 1;
  int ret = 0;
  while (n){
    ret++;
    n /= 10;
  }
  return ret;
}

int main()
{
  FastIO();
  //int tc, ca = 0;
  int n;
  cin >> n;
  vector <Long> v(n);
  for (int i = 0;i < n;i++){
    cin >> v[i];
  }
  Long prod = 1;
  Long lim = 1e18;
  for (int i = 0;i < n;i++){
    if (d(v[i])+d(prod) > 20){
      prod = -1;
      break;
    }
    prod *= v[i];
    if (prod > lim){
      //cout << "test" << '\n';
      prod = -1;
      break;
    }
  }
  cout << prod << '\n';
  return 0;
}