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;
  inline void pisz(int n) { printf("%d\n", n); }
  const int dr[4] = {-1, 0, 1, 0};
  const int dc[4] = {0, -1, 0, 1};
  const int drr[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
  const int dcc[8] = {0, -1, -1, -1, 0, 1, 1, 1};
  template <typename T, typename TT>
  ostream& operator<<(ostream& s, pair<T, TT> t) {
    return s << "(" << t.first << "," << t.second << ")";
  }
  template <typename T>
  ostream& operator<<(ostream& s, vector<T> t) {
    for (int i = 0; i < ((int)((t).size())); i++) s << t[i] << " ";
    return s;
  }
  class UnionFind {
   private:
    vector<int> rank, p;
  
   public:
    UnionFind(int n) {
      for (int i = 0; i < n; i++) p.push_back(i);
      rank.assign(n, 0);
    }
    int findSet(int s) { return (p[s] == s) ? s : (p[s] = findSet(p[s])); }
    bool isSameSet(int i, int j) { return findSet(i) == findSet(j); }
    void unionSet(int i, int j) {
      if (isSameSet(i, j)) return;
      int x = findSet(i), y = findSet(j);
      if (rank[x] > rank[y])
        p[y] = x;
      else {
        p[x] = y;
        if (rank[x] == rank[y]) rank[y]++;
      }
    }
  };
  bool fcomp(double A, double B) {
    double EPSILON = numeric_limits<double>::epsilon();
    double diff = A - B;
    return fabs(diff) < EPSILON;
  }
  double x_y(int x1, int y1, int x2, int y2) {
    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
  }
  int main() {
    int(testow);
    scanf("%d", &(testow));
    while (testow--) {
      long long ct[2];
      memset(ct, 0, sizeof(ct));
      int(n);
      scanf("%d", &(n));
      int dt[n];
      for (int i = 0; i < n; i++) {
        scanf("%d", &dt[i]);
        if (dt[i] == 0 || dt[i] == 1) ct[dt[i]]++;
      }
      long long mul = 1;
      for (int i = 1; i <= ct[0]; i++) mul *= 2;
      cout << ct[1] * mul << endl;
    }
    return 0;
  }
  