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: Accepted)

import java.util.Arrays;
import java.util.Scanner;
 
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        //入力
      	int n = sc.nextInt();
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
        }
 		//0を見つけたいので昇順にソート
        Arrays.sort(a);
 		//10の18乗
        final long K = (long) Math.pow(10, 18);
 		//0があるとき
        if (a[0] == 0) {
            System.out.println(0);
            return;
        }
 
        long result = 1;
      	//取り出した値を格納する変数:順に取り出したい配列
        for (long l : a) {
            if (result > K / l) {
                System.out.println(-1);
                return;
            }
          //取り出した値を掛ける
            result *= l;
        }
        System.out.println(result);
 
    }
}