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.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();
  }
}

