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<stdio.h>
#include<math.h>
#include<string.h>
 
int keta(long long int xx)
{
	int yy = 0;
	while (1) {
		xx = xx / 10;
		yy++;
		if (xx < 10) {
			break;
		}
	}
	return yy;
}
 
int ke(long long int xx, long long int yy) {
	int xxx = 0;
	while (1) {
		xx = xx / 10;
		xxx++;
		if (xx < 10) {
			break;
		}
	}
	int yyy = 0;
	while (1) {
		yy = yy / 10;
		yyy++;
		if (yy < 10) {
			break;
		}
	}
	if (xx*yy > 11) {
		return 1;
	}
	return 0;
}
 
 
int main() {
	int i, j, k;
	int  n;
	long long int f = 1, m = 0;
	int g = 0;
	int x[110] = {0};
	int a, b, c, d;
 
 
	scanf("%d", &n);
 
	for (i = 0;i < n;i++) {
		scanf("%lld", &m);
 
		if (keta(f) + keta(m) > 18) {
			f = -1;
			break;
		}
 
		f = f * m;
 
		if (keta(f) > 18) {
			f = -1;
			break;
		}
	}
 
	printf("%lld", f);
 
	return 0;
}