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>
  long long q_pow(long long x, long long n, long long mod) {
    x %= mod;
    long long ret = 1;
    while (n) {
      if (n & 1) ret = x * ret % mod;
      x = x * x % mod;
      n >>= 1;
    }
    return ret;
  }
  long long getbit(long long now) {
    int cnt = 0;
    while (now) {
      cnt++;
      now -= (now) & (-now);
    }
    return cnt;
  }
  int dx[] = {1, 0};
  int dy[] = {0, 1};
  using namespace std;
  void dbg_out() { cout << "\n"; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cout << ' ' << H;
    dbg_out(T...);
  }
  const long long maxn = 1e6 + 5;
  const long long maxm = 250 + 10;
  const long long INF = 0x3f3f3f3f;
  const long long IINF = 0x3f3f3f3f3f3f3f3f;
  const long long mod = 998244353;
  void add(long long &x, long long y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  void mul(long long &x, long long y) {
    x -= y;
    if (x < 0) x += mod;
  }
  long long a[maxn];
  map<long long, long long> g;
  map<long long, long long> num;
  void solve() {
    int n;
    cin >> n;
    long long cnt0 = 0;
    long long cnt1 = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i] == 0) cnt0++;
      if (a[i] == 1) cnt1++;
    }
    cout << cnt1 * (1LL << cnt0) << "\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  