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;
  struct _IO {
    _IO() {
      ios::sync_with_stdio(0);
      cin.tie(0);
      cout.tie(0);
    }
  } _io;
  const long long maxn = 1e6 + 10, mod = 1e9 + 7;
  template <typename... T>
  long long madd(T... args) {
    long long res = 0;
    for (auto i : {args...}) {
      res = (res + ((i % mod + mod) % mod)) % mod;
    }
    return res % mod;
  }
  template <typename... T>
  long long mmul(T... args) {
    long long res = 1;
    for (auto i : {args...}) {
      res = (res * ((i % mod + mod) % mod)) % mod;
    }
    return res % mod;
  }
  long long qpow(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res *= a;
      a *= a;
      b >>= 1;
    }
    return res;
  }
  long long mpow(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b >>= 1;
    }
    return res % mod;
  }
  long long _inv(long long x) { return mpow(x, mod - 2); }
  void solve() {
    long long cnt0 = 0, cnt1 = 0;
    long long n;
    cin >> n;
    long long t;
    while (n--) {
      cin >> t;
      if (t == 1)
        ++cnt1;
      else if (t == 0)
        ++cnt0;
    }
    cout << cnt1 * qpow(2ll, cnt0) << '\n';
  }
  int main() {
    long long T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  