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: #include <bits/stdc++.h>
  using namespace std;
  const int M = 1e9 + 7;
  long long fastpower(long long x, long long n, long long M) {
    if (n == 0)
      return 1;
    else if (n % 2 == 0)
      return fastpower((x * x) % M, n / 2, M);
    else
      return (x * fastpower((x * x) % M, (n - 1) / 2, M)) % M;
  }
  long long GCD(long long A, long long B) {
    if (B == 0)
      return A;
    else
      return GCD(B, A % B);
  }
  bool vowl(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
  }
  long long modInverse(long long A, long long M) {
    return fastpower(A, M - 2, M);
  }
  void sieve(long long N) {
    bool isPrime[N + 1];
    for (long long i = 0; i <= N; ++i) {
      isPrime[i] = true;
    }
    isPrime[0] = false;
    isPrime[1] = false;
    for (long long i = 2; i * i <= N; ++i) {
      if (isPrime[i] == true) {
        for (long long j = i * i; j <= N; j += i) isPrime[j] = false;
      }
    }
  }
  vector<long long> factorize(long long n) {
    vector<long long> res;
    for (long long i = 2; i * i <= n; ++i) {
      while (n % i == 0) {
        res.push_back(i);
        n /= i;
      }
    }
    if (n != 1) {
      res.push_back(n);
    }
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long arr[n];
      for (long long i = 0; i < n; i++) cin >> arr[i];
      long long one = 0, zero = 0;
      for (long long i = 0; i < n; i++) {
        if (arr[i] == 1) one++;
        if (arr[i] == 0) zero++;
      }
      long long ans = ((long long)1 << zero);
      ans *= (long long)one;
      cout << ans << endl;
    }
    return 0;
  }
  