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.File;
  import java.lang.invoke.MethodHandles;
  import java.util.Scanner;
  
  public class R750B {
  
    static Scanner getInputScanner() {
      try {
        final String USER_DIR = System.getProperty("user.dir");
        final String CNAME = MethodHandles.lookup().lookupClass().getSimpleName();
        File fin = new File(USER_DIR + "/io/" + CNAME + ".in");
        if (fin.exists()) {
          return new Scanner(fin);
        }
      } catch (Exception e) {
      }
      return new Scanner(System.in);
    }
  
    static long solve(int n, int[] a) {
      int m0 = 0;
      int m1 = 0;
      for (int v : a) {
        if (v == 0) {
          m0++;
        } else if (v == 1) {
          m1++;
        }
      }
      if (m1 == 0) {
        return 0;
      }
      // Exclude one 1 and any combination of 0
      return 1L * m1 * (1L << m0);
    }
  
    public static void main(String[] args) {
      Scanner in = getInputScanner();
      int T = in.nextInt();
      for (int t = 1; t <= T; t++) {
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
          a[i] = in.nextInt();
        }
        System.out.format("%d\n", solve(n, a));
      }
      in.close();
    }
  }
  
  