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.util.*;
  
  public class Question1 {
      static Scanner in = new Scanner(System.in);
      public static void main(String[] args) {
          int t = in.nextInt();
          while(t-->0){
              int n = in.nextInt();
  //            int[] arr = new int[n];
              int set = 0,z = 0,a;
              for(int i = 0;i<n;i++){
                  a = in.nextInt();
                  if(a==1){
                      set++;
                  }else if (a==0){
                      z++;
                  }
  //                arr[i] = in.nextInt();
              }
              if(z==0){
                  System.out.println(set);
              }else if(set==0){
                  System.out.println(0);
              }else{
                  long ans = set + ((long)Math.pow(2,z)-1)*set;
                  System.out.println(ans);
              }
          }
      }
  }
  