description: 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).
solution: import java.io.*;
  import java.util.StringTokenizer;
  
  public class Main {
  	
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter pw = new PrintWriter(System.out);
  	
  	public static void main(String[] args) throws IOException {
  		StringTokenizer st;
  		short t = Short.parseShort(br.readLine());
  		byte n, ones, zeros;
  		
  		for (short i=0; i<t; i++) {
  			ones = 0; zeros = 0;
  			n = Byte.parseByte(br.readLine());
  			st = new StringTokenizer(br.readLine());
  			for (byte ii=0; ii<n; ii++) {
  				int number = Integer.parseInt(st.nextToken());
  				if (number == 1) {
  					ones++;
  				} else if (number == 0) {
  					zeros++;
  				}
  			}
  			
  			pw.println((long)(Math.pow(2, zeros) * ones));
  		}
  		
  		pw.close();
  	}
  }
  