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>
  #pragma GCC optimize("O3", "unroll-loops")
  #pragma GCC target("avx2")
  using namespace std;
  long long int gcd(long long int a, long long int b) {
    if (a == 0)
      return b;
    else if (b == 0)
      return a;
    if (a > b)
      return gcd(a % b, b);
    else
      return gcd(b % a, a);
  }
  long long int power(long long int x, long long int y, long long int p) {
    long long int res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long int modInv(long long int a, long long int b) {
    return power(a, b - 2, b);
  }
  long long int add(long long int a, long long int b, long long int MOD) {
    long long int r = a + b;
    if (r >= MOD) r -= MOD;
    if (r < 0) r += MOD;
    return r;
  }
  long long int mul(long long int a, long long int b, long long int MOD) {
    long long int r = (a % MOD) * (b % MOD);
    r %= MOD;
    if (r < 0) r += MOD;
    return r;
  }
  long long int Min3(long long int a, long long int b, long long int c) {
    if (a < b)
      return (a < c ? a : c);
    else
      return (b < c ? b : c);
  }
  long long int power(long long int x, long long int y) {
    long long int res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x);
      y = y >> 1;
      x = (x * x);
    }
    return res;
  }
  void solve() {
    long long int n, c0 = 0, c1 = 0;
    cin >> n;
    vector<long long int> v(n);
    for (long long int i = 0; i < n; i++) {
      cin >> v[i];
      if (v[i] == 0)
        c0++;
      else if (v[i] == 1)
        c1++;
    }
    long long int ans = c1 * (power(2, c0));
    cout << ans << "\n";
  }
  signed main() {
    clock_t start, end;
    start = clock();
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int tcs = 1;
    cin >> tcs;
    while (tcs--) solve();
    end = clock();
    double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
    cerr << time_taken << " sec";
    return 0;
  }
  