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.math.BigInteger;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long[] numberList = new long[n];
		BigInteger answer = BigInteger.ONE;
		for(int i = 0;i < n;i++) {
			long number = sc.nextLong();
			if(number == 0) {
				System.out.println(0);
				return;
			}
			numberList[i] = number;

		}

		for(int i =0;i < n;i++) {

			answer = answer.multiply(new BigInteger(String.valueOf(numberList[i])));
			if(answer.compareTo(new BigInteger(String.valueOf((long)Math.pow(10, 18)))) >=1) {
				System.out.println(-1);
				return;
			}
		}

		System.out.println(answer);
	}

}