#include <bits/stdc++.h>
using namespace std;
int n;
long long a[100010], mx, f[70][100010], b[100010], id[100010], s1[100010],
    s0[100010];
int cmp(int x, int y) { return b[x] < b[y]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) a[i] = mx - a[i];
  memset(f, 63, sizeof(f));
  for (int i = 1; i <= n; i++) id[i] = i;
  f[0][0] = 0;
  for (long long i = 0; i < 63; i++) {
    for (long long j = 1; j <= n; j++)
      b[id[j]] = ((1ll << (i)) - 1ll) & a[id[j]];
    sort(id + 1, id + n + 1, cmp);
    memset(s1, 0, sizeof(s1));
    memset(s0, 0, sizeof(s0));
    for (int j = 1; j <= n; j++) {
      s1[j] = s1[j - 1];
      s0[j] = s0[j - 1];
      if (a[id[j]] & (1ll << i))
        s1[j]++;
      else
        s0[j]++;
    }
    for (long long j = 0; j <= n; j++)
      for (long long k = 0; k < 2; k++) {
        if (!k) {
          long long nxt = s1[n] - s1[n - j], va = s0[n] - s0[n - j] + s1[n - j];
          f[i + 1][nxt] = min(f[i + 1][nxt], f[i][j] + va);
        } else {
          long long nxt = n - s0[n - j], va = s1[n] - s1[n - j] + s0[n - j];
          f[i + 1][nxt] = min(f[i + 1][nxt], f[i][j] + va);
        }
      }
  }
  printf("%I64d", f[63][0]);
}

### Prompt

In cpp, your task is to solve 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 n;
long long a[100010], mx, f[70][100010], b[100010], id[100010], s1[100010],
    s0[100010];
int cmp(int x, int y) { return b[x] < b[y]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) a[i] = mx - a[i];
  memset(f, 63, sizeof(f));
  for (int i = 1; i <= n; i++) id[i] = i;
  f[0][0] = 0;
  for (long long i = 0; i < 63; i++) {
    for (long long j = 1; j <= n; j++)
      b[id[j]] = ((1ll << (i)) - 1ll) & a[id[j]];
    sort(id + 1, id + n + 1, cmp);
    memset(s1, 0, sizeof(s1));
    memset(s0, 0, sizeof(s0));
    for (int j = 1; j <= n; j++) {
      s1[j] = s1[j - 1];
      s0[j] = s0[j - 1];
      if (a[id[j]] & (1ll << i))
        s1[j]++;
      else
        s0[j]++;
    }
    for (long long j = 0; j <= n; j++)
      for (long long k = 0; k < 2; k++) {
        if (!k) {
          long long nxt = s1[n] - s1[n - j], va = s0[n] - s0[n - j] + s1[n - j];
          f[i + 1][nxt] = min(f[i + 1][nxt], f[i][j] + va);
        } else {
          long long nxt = n - s0[n - j], va = s1[n] - s1[n - j] + s0[n - j];
          f[i + 1][nxt] = min(f[i + 1][nxt], f[i][j] + va);
        }
      }
  }
  printf("%I64d", f[63][0]);
}

```