#include <bits/stdc++.h>
using namespace std;
pair<long long, int> res[60][100010];
long long a[100010];
int dp[60][100010];
int tot[60];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 60; j++) {
      if ((a[i] >> j) & 1) tot[j]++;
      long long first = (a[i] >> j + 1) << j + 1;
      res[j][i - 1] = pair<long long, int>(a[i] - first, i);
    }
  }
  for (int j = 0; j < 60; j++) {
    sort(res[j], res[j] + n), reverse(res[j], res[j] + n);
  }
  for (int i = 0; i < 60; i++)
    for (int j = 0; j < 100010; j++) dp[i][j] = 0x3f3f3f3f;
  dp[0][tot[0]] = n - tot[0];
  dp[0][0] = tot[0];
  for (int i = 0; i < 60 - 1; i++) {
    int cnt = 0;
    for (int j = 0; j <= n; j++) {
      if (j and (a[res[i][j - 1].second] >> i + 1) % 2) cnt++;
      if (dp[i][j] == 0x3f3f3f3f) continue;
      dp[i + 1][cnt] =
          min(dp[i + 1][cnt], dp[i][j] + tot[i + 1] - cnt + (j - cnt));
      dp[i + 1][tot[i + 1] + j - cnt] =
          min(dp[i + 1][tot[i + 1] + j - cnt],
              dp[i][j] + n - (tot[i + 1] - cnt + (j - cnt)));
    }
  }
  cout << dp[60 - 1][0] << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
pair<long long, int> res[60][100010];
long long a[100010];
int dp[60][100010];
int tot[60];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 60; j++) {
      if ((a[i] >> j) & 1) tot[j]++;
      long long first = (a[i] >> j + 1) << j + 1;
      res[j][i - 1] = pair<long long, int>(a[i] - first, i);
    }
  }
  for (int j = 0; j < 60; j++) {
    sort(res[j], res[j] + n), reverse(res[j], res[j] + n);
  }
  for (int i = 0; i < 60; i++)
    for (int j = 0; j < 100010; j++) dp[i][j] = 0x3f3f3f3f;
  dp[0][tot[0]] = n - tot[0];
  dp[0][0] = tot[0];
  for (int i = 0; i < 60 - 1; i++) {
    int cnt = 0;
    for (int j = 0; j <= n; j++) {
      if (j and (a[res[i][j - 1].second] >> i + 1) % 2) cnt++;
      if (dp[i][j] == 0x3f3f3f3f) continue;
      dp[i + 1][cnt] =
          min(dp[i + 1][cnt], dp[i][j] + tot[i + 1] - cnt + (j - cnt));
      dp[i + 1][tot[i + 1] + j - cnt] =
          min(dp[i + 1][tot[i + 1] + j - cnt],
              dp[i][j] + n - (tot[i + 1] - cnt + (j - cnt)));
    }
  }
  cout << dp[60 - 1][0] << endl;
  return 0;
}

```