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)

int main(){
  int num, i;
  int cnt = 0;
  long int* a;
  long int b;
  long int ans = 1;
  
  scanf("%d", &num);
  a = (long int *)malloc(sizeof(long int)*num);
  
  for(i = 0; i < num; i++){
    scanf("%ld ", &a[i]);
    b = a[i];
    if(a[i] == 0){
      printf("%d", 0);
      return 0;
    }
    if(b >= 10){
    	while(b >= 10) {
      		b /= 10;
      		cnt++;
    	}
    }
  }
  
  for(i = 0; i < num ; i++){
  	ans *= a[i];
    if((ans > 1000000000000000000) || (ans < 0)){
      break;
    }
  }
  
  if((cnt > 18) || (ans > 1000000000000000000) || (ans < 0)) {
    ans = -1;
    printf("%ld", ans);
  }
  else
  {
    printf("%ld", ans);
  }
  
  return 0;
}