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;
  template <class T>
  void read(T& x) {
    cin >> x;
  }
  template <class H, class... T>
  void read(H& h, T&... t) {
    read(h);
    read(t...);
  }
  template <class A>
  void read(A* x, int size) {
    for (int i = (0); (1) > 0 ? i < (size) : i > (size); i += (1)) read(x[i]);
  }
  template <class A>
  void read(vector<A>& x) {
    for (auto& a : x) read(a);
  }
  template <class T>
  void print(T& x) {
    cout << x << ' ';
  }
  template <class H, class... T>
  void print(H h, T... t) {
    print(h);
    print(t...);
  }
  template <class A>
  void print(A* x, int size) {
    for (int i = (0); (1) > 0 ? i < (size) : i > (size); i += (1)) print(x[i]);
    cout << '\n';
  }
  template <class A>
  void print(vector<A>& x) {
    for (auto& a : x) print(a);
  }
  template <class H, class... T>
  void debug(H h, T... t) {
    print(h, t...);
    cout << endl;
  }
  const int d4i[4] = {-1, 0, 1, 0}, d4j[4] = {0, 1, 0, -1};
  const int d8i[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
            d8j[8] = {0, 1, 1, 1, 0, -1, -1, -1};
  const int N = 200010, mod = 1e9 + 7;
  void solve() {
    long long sum = 0, one = 0, zero = 0;
    int n, val;
    read(n);
    for (int i = (0); (1) > 0 ? i < (n) : i > (n); i += (1)) {
      read(val);
      sum += val;
      one += (val == 1);
      zero += (val == 0);
    }
    if (sum == 0)
      cout << 0 << '\n';
    else if (sum == 1)
      cout << (1LL << zero) << '\n';
    else
      cout << (1LL << zero) * one << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int _t;
    cin >> _t;
    while (_t--) solve();
  }
  