#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 100005;
ll a[N];
int n, f[66][N], id[N];
int sum[2][N];
int main() {
  ios::sync_with_stdio(false);
  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], id[i] = i;
  memset(f, 63, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k <= 60; k++) {
    sort(id + 1, id + n + 1, [&](int x, int y) {
      return (a[x] & ((1ll << k) - 1)) < (a[y] & ((1ll << k) - 1));
    });
    for (int i = 1; i <= n; i++) {
      sum[0][i] = sum[0][i - 1];
      sum[1][i] = sum[1][i - 1];
      ++sum[(a[id[i]] >> k) & 1][i];
    }
    for (int i = 0; i <= n; i++) {
      int ww, tt;
      ww = sum[1][n - i] + sum[0][n] - sum[0][n - i];
      tt = sum[1][n] - sum[1][n - i];
      f[k + 1][tt] = min(f[k + 1][tt], f[k][i] + ww);
      ww = sum[0][n - i] + sum[1][n] - sum[1][n - i];
      tt = n - sum[0][n - i];
      f[k + 1][tt] = min(f[k + 1][tt], f[k][i] + ww);
    }
  }
  cout << f[61][0] << endl;
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
const int N = 100005;
ll a[N];
int n, f[66][N], id[N];
int sum[2][N];
int main() {
  ios::sync_with_stdio(false);
  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], id[i] = i;
  memset(f, 63, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k <= 60; k++) {
    sort(id + 1, id + n + 1, [&](int x, int y) {
      return (a[x] & ((1ll << k) - 1)) < (a[y] & ((1ll << k) - 1));
    });
    for (int i = 1; i <= n; i++) {
      sum[0][i] = sum[0][i - 1];
      sum[1][i] = sum[1][i - 1];
      ++sum[(a[id[i]] >> k) & 1][i];
    }
    for (int i = 0; i <= n; i++) {
      int ww, tt;
      ww = sum[1][n - i] + sum[0][n] - sum[0][n - i];
      tt = sum[1][n] - sum[1][n - i];
      f[k + 1][tt] = min(f[k + 1][tt], f[k][i] + ww);
      ww = sum[0][n - i] + sum[1][n] - sum[1][n - i];
      tt = n - sum[0][n - i];
      f[k + 1][tt] = min(f[k + 1][tt], f[k][i] + ww);
    }
  }
  cout << f[61][0] << endl;
  return 0;
}

```