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.*;
  
  public class Main {
      public static void main(String[] args) throws IOException {
          int t = nextInt();
          while (t-- != 0) {
              int n = nextInt();
              long[] a = new long[n];
              int count0 = 0;
              int count1 = 0;
              for (int i = 0; i < n; i++) {
                  a[i] = nextLong();
                  if (a[i] == 0) count0++;
                  else if (a[i] == 1) count1++;
              }
              long ans = 0;
              for (int i = 0; i <= count0; i++) {
                  long y = 1;
                  for (int j = 1; j <= i; j++) {
                      y *= (count0 - j + 1);
                      y/=j;
                  }
                  ans += y * count1;
              }
              out.println(ans);
          }
          out.close();
      }
  
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter out = new PrintWriter(System.out);
      static StringTokenizer in = new StringTokenizer("");
  
  
      public static boolean hasNext() throws IOException {
          if (in.hasMoreTokens()) return true;
          String s;
          while ((s = br.readLine()) != null) {
              in = new StringTokenizer(s);
              if (in.hasMoreTokens()) return true;
          }
          return false;
      }
  
      public static String nextToken() throws IOException {
          while (!in.hasMoreTokens()) {
              in = new StringTokenizer(br.readLine());
          }
          return in.nextToken();
      }
  
      public static int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      public static long nextLong() throws IOException {
          return Long.parseLong(nextToken());
      }
  
      public static double nextDouble() throws IOException {
          return Double.parseDouble(nextToken());
      }
  }
  
  class Ayo {
      int v;
      long w;
  
      public Ayo(int v, long w) {
          this.v = v;
          this.w = w;
      }
  }
  
  class Ayo1 implements Comparable<Ayo1> {
      long w;
      int v;
  
      public Ayo1(long w, int v) {
          this.w = w;
          this.v = v;
      }
  
      @Override
      public int compareTo(Ayo1 o) {
          if (o.v == v && o.w == w) return 0;
          else {
              if (w < o.w) return -1;
              else if (w > o.w) return 1;
              else {
                  return Integer.compare(v, o.v);
              }
          }
      }
  }