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.*;
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);
            }
        }
    }
}