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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.math.BigDecimal;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        BigDecimal[] nums = new BigDecimal[N];
        for (int i = 0; i < N; i++) {
            nums[i] = new BigDecimal(sc.next());
        }
        Arrays.sort(nums);
        BigDecimal ans = nums[0];
        if (ans.longValue() == 0) {
            System.out.println(0);
        } else {
            boolean over = false;
            for (int i = 1; i < N; i++) {
                ans = ans.multiply(nums[i]);
                if (1000000000000000000L < ans.longValue()) {
                    System.out.println(-1);
                    over = true;
                    break;
                }
            }
            if (!over) {
                System.out.println(ans.longValue());
            }
        }
    }
}