There are n positive integers a_1, a_2, ..., a_n. For the one move you can choose any even value c and divide by two all elements that equal c.

For example, if a=[6,8,12,6,3,12] and you choose c=6, and a is transformed into a=[3,8,12,3,3,12] after the move.

You need to find the minimal number of moves for transforming a to an array of only odd integers (each element shouldn't be divisible by 2).

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.

The first line of a test case contains n (1 ≤ n ≤ 2⋅10^5) — the number of integers in the sequence a. The second line contains positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

The sum of n for all test cases in the input doesn't exceed 2⋅10^5.

Output

For t test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by 2).

Example

Input


4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7


Output


4
10
4
0

Note

In the first test case of the example, the optimal sequence of moves can be as follows:

  * before making moves a=[40, 6, 40, 3, 20, 1]; 
  * choose c=6; 
  * now a=[40, 3, 40, 3, 20, 1]; 
  * choose c=40; 
  * now a=[20, 3, 20, 3, 20, 1]; 
  * choose c=20; 
  * now a=[10, 3, 10, 3, 10, 1]; 
  * choose c=10; 
  * now a=[5, 3, 5, 3, 5, 1] — all numbers are odd. 



Thus, all numbers became odd after 4 moves. In 3 or fewer moves, you cannot make them all odd.
#include <bits/stdc++.h>
auto clk = clock();
const int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
const int dx8[] = {-1, -1, -1, 0, 1, 1, 1, 0},
          dy8[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long Gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return Gcd(b, a % b);
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void printvector(std::vector<long long> v) {
  for (int i = 0; i < v.size(); ++i) {
    cerr << v[i] << " ";
  }
  cerr << "" << endl;
}
void printarray(long long a[], long long n) {
  for (long long i = 0; i < n; ++i) {
    cerr << a[i] << " ";
  }
  cerr << "" << endl;
}
long long findlcm(vector<long long> arr, long long n) {
  long long ans = arr[0];
  for (int i = 1; i < n; i++) ans = (((arr[i] * ans)) / (Gcd(arr[i], ans)));
  return ans;
}
long long power(long long x, long long y, long long p) {
  long long 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 lowerbound(long long k, vector<long long> v) {
  long long l = 0;
  long long size = v.size();
  long long r = size - 1;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] < k)
      l = mid + 1;
    else {
      r = mid - 1;
    }
  }
  return l;
}
long long upperbound(long long k, vector<long long> v) {
  long long l = 0;
  long long size = v.size();
  long long r = size - 1;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] > k)
      r = mid - 1;
    else {
      l = mid + 1;
    }
  }
  return l;
}
long long spf[100010];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < 100010; i++) spf[i] = i;
  for (int i = 4; i < 100010; i += 2) spf[i] = 2;
  for (int i = 3; i * i < 100010; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < 100010; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long n, m;
vector<long long> shortPath(vector<pair<long long, long long>> v[],
                            long long src, long long V = 0) {
  set<pair<long long, long long>> setds;
  long long sizee = V;
  long long IN = LLONG_MAX;
  vector<long long> dist(sizee, IN);
  vector<bool> vis(sizee, false);
  setds.insert(make_pair(0, src));
  dist[src] = 0;
  while (!setds.empty()) {
    pair<long long, long long> tmp = *(setds.begin());
    setds.erase(setds.begin());
    long long u = tmp.second;
    if (vis[u]) continue;
    vis[u] = true;
    vector<pair<long long, long long>>::iterator i;
    for (auto i = v[u].begin(); i != v[u].end(); ++i) {
      long long v = (*i).first;
      long long weight = (*i).second;
      if (dist[v] > dist[u] + weight) {
        if (dist[v] != LLONG_MAX)
          setds.erase(setds.find(make_pair(dist[v], v)));
        dist[v] = dist[u] + weight;
        setds.insert(make_pair(dist[v], v));
      }
    }
  }
  return dist;
}
long long root(long long a[], long long r) {
  if (a[r] == r) {
    return r;
  } else {
    long long x = root(a, a[r]);
    a[r] = x;
    return x;
  }
}
void unio(long long a[], long long size[], long long x, long long y) {
  x = root(a, x);
  y = root(a, y);
  if (x == y) {
    return;
  }
  if (size[x] >= size[y]) {
    a[y] = x;
    size[x] += size[y];
  } else {
    a[x] = y;
    size[y] += size[x];
  }
}
void neuer(long long txt) {
  cin >> n;
  vector<long long> ar(n);
  set<long long> st;
  for (int i = 0, _n = (n); i < _n; i++) {
    cin >> ar[i];
    st.insert(ar[i]);
  }
  long long op = 0;
  while (st.size()) {
    auto it = st.end();
    it--;
    long long x = *it;
    st.erase(*it);
    if (x & 1) {
    } else {
      st.insert(x / 2);
      op++;
    }
  }
  cout << op << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long x68 = 1;
  cin >> x68;
  for (long long vi = 1; vi <= x68; vi++) {
    neuer(vi);
  }
}
