Luntik came out for a morning stroll and found an array a of length n. He calculated the sum s of the elements of the array (s= ∑_{i=1}^{n} a_i). Luntik calls a subsequence of the array a nearly full if the sum of the numbers in that subsequence is equal to s-1.

Luntik really wants to know the number of nearly full subsequences of the array a. But he needs to come home so he asks you to solve that problem!

A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The next 2 ⋅ t lines contain descriptions of test cases. The description of each test case consists of two lines.

The first line of each test case contains a single integer n (1 ≤ n ≤ 60) — the length of the array.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — the elements of the array a.

Output

For each test case print the number of nearly full subsequences of the array.

Example

Input


5
5
1 2 3 4 5
2
1000 1000
2
1 0
5
3 0 2 1 1
5
2 1 0 3 0


Output


1
0
2
4
4

Note

In the first test case, s=1+2+3+4+5=15, only (2,3,4,5) is a nearly full subsequence among all subsequences, the sum in it is equal to 2+3+4+5=14=15-1.

In the second test case, there are no nearly full subsequences.

In the third test case, s=1+0=1, the nearly full subsequences are (0) and () (the sum of an empty subsequence is 0).
import java.util.*;

public class Main {

    static boolean isPrime(int n)
    {
        // Corner case
        if (n <= 1)
            return false;
  
        // Check from 2 to n-1
        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;
  
        return true;
    }
    
    
	public static void main (String[] args){
		Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        

        while(T>0)
        {
        int  n = sc.nextInt();
        long arr[] = new long [n];
        
        int a = 0;
        int b = 0;
        
        
        for(int i=0;i<n;i++)
        {
            arr[i] = sc.nextLong();
            if(arr[i]==0)
            a++;
            
            if(arr[i]==1)
            b++;
        }   
        
       
            double k = 0;
            double l = Math.pow(2,a);
            
            k = b*l;
            
            long j = (long)k;
            System.out.println(j);
            
        
            
        
        
        T--;
        }
    }
}
