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.*; 
import java.io.*;

public class Main {
    static StringTokenizer st;
    static BufferedReader br;  
    static PrintWriter out;
 
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));

        int N = nextInt();
        long[] input = new long[N];

        for (int i=0; i<N; i++) {
            input[i] = nextLong();
            if(input[i]==0) {
                System.out.println(0);
                return;
            }
        }

        long ans =input[0];

        for (int i=1; i<N; i++) {
            // If you directly perform multiplication on 64-bit integers, it causes overflow
            // if((Math.log10(ans) + Math.log10(input[i]) >= 18) && ans * input[i] != 1e18){
            //if(ans> 1e18/ input[i]){ 
            // > not working, need use <=, because 1e18/ input[i] will lose 0.000001
            if(!(ans<= (long)(1e18)/ input[i])){ 
                System.out.println(-1);
                return;
            }
            ans = ans * input[i];
        }

        System.out.println(ans);
 
    }
    
      
    public static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    public static long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
}