#include <bits/stdc++.h>
using namespace std;
int main() {
  do {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  } while (false);
  int n;
  cin >> n;
  vector<long long> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  const int LOG = 60;
  vector<vector<long long> > appear(LOG, vector<long long>(n));
  for (int i = 1; i < LOG; i++) {
    for (int j = 0; j < n; j++) {
      appear[i][j] = arr[j] & ((1ll << i) - 1);
    }
    sort(appear[i].begin(), appear[i].end());
  }
  vector<vector<long long> > dp(LOG, vector<long long>(n + 1, LLONG_MAX));
  dp[0][n] = 0;
  for (int i = 0; i + 1 < LOG; i++) {
    auto calculate = [&](long long value) {
      return upper_bound(appear[i + 1].begin(), appear[i + 1].end(), value) -
             appear[i + 1].begin();
    };
    for (int j = 0; j <= n; j++) {
      if (dp[i][j] != LLONG_MAX) {
        long long current = (j ? appear[i][j - 1] : 0);
        for (int value = 0; value < 2; value++) {
          long long next = current + (1ll << i) * value;
          int k = calculate(next);
          int between = calculate(current + (1ll << i)) - calculate(current);
          dp[i + 1][k] =
              min(dp[i + 1][k], dp[i][j] + (!value ? between : n - between));
        }
      }
    }
  }
  cout << dp[LOG - 1][n] << endl;
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

You are given n numbers a_1, a_2, ..., a_n. In one operation we can add to any one of those numbers a nonnegative integer power of 2.

What is the smallest number of operations we need to perform to make all n numbers equal? It can be proved that under given constraints it doesn't exceed 10^{18}.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^{17}).

Output

Output exactly one integer — the smallest number of operations we need to perform to make all n numbers equal.

Examples

Input


4
228 228 228 228


Output


0

Input


3
2 2 8


Output


3

Note

In the first example, all numbers are already equal. So the needed number of operation is 0.

In the second example, we can apply the operation 3 times: add 8 to first 2, add 8 to second 2, add 2 to 8, making all numbers equal to 10. It can be proved that we can't make all numbers equal in less than 3 operations.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  do {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  } while (false);
  int n;
  cin >> n;
  vector<long long> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  const int LOG = 60;
  vector<vector<long long> > appear(LOG, vector<long long>(n));
  for (int i = 1; i < LOG; i++) {
    for (int j = 0; j < n; j++) {
      appear[i][j] = arr[j] & ((1ll << i) - 1);
    }
    sort(appear[i].begin(), appear[i].end());
  }
  vector<vector<long long> > dp(LOG, vector<long long>(n + 1, LLONG_MAX));
  dp[0][n] = 0;
  for (int i = 0; i + 1 < LOG; i++) {
    auto calculate = [&](long long value) {
      return upper_bound(appear[i + 1].begin(), appear[i + 1].end(), value) -
             appear[i + 1].begin();
    };
    for (int j = 0; j <= n; j++) {
      if (dp[i][j] != LLONG_MAX) {
        long long current = (j ? appear[i][j - 1] : 0);
        for (int value = 0; value < 2; value++) {
          long long next = current + (1ll << i) * value;
          int k = calculate(next);
          int between = calculate(current + (1ll << i)) - calculate(current);
          dp[i + 1][k] =
              min(dp[i + 1][k], dp[i][j] + (!value ? between : n - between));
        }
      }
    }
  }
  cout << dp[LOG - 1][n] << endl;
  return 0;
}

```