#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, f[2][N];
long long a[N], mx, v1[N], v2[N];
void upd(int& x, int y) { x = (x < y ? x : y); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]), mx = max(mx, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = mx - a[i];
  memset(f[1], 0x3f, sizeof(int) * (n + 1));
  f[1][0] = 0;
  for (int i = 0; i < 64; ++i) {
    memset(f[i & 1], 0x3f, sizeof(int) * (n + 1));
    int c[2] = {0}, t = 0;
    for (int j = 1; j <= n; ++j) ++c[a[j] >> i & 1];
    upd(f[i & 1][0], f[~i & 1][0] + c[1]);
    upd(f[i & 1][c[1]], f[~i & 1][0] + c[0]);
    for (int j = 1; j <= n; ++j) {
      if (a[j] >> i & 1)
        --c[1], ++c[0], ++t;
      else
        --c[0], ++c[1];
      upd(f[i & 1][t], f[~i & 1][j] + c[1]);
      upd(f[i & 1][t + c[1]], f[~i & 1][j] + c[0]);
    }
    int id1 = 0, id2 = 0, id = 0;
    for (int j = 1; j <= n; ++j)
      (a[j] >> i & 1) ? v1[++id1] = a[j] : v2[++id2] = a[j];
    for (int j = 1; j <= id1; ++j) a[++id] = v1[j];
    for (int j = 1; j <= id2; ++j) a[++id] = v2[j];
  }
  printf("%d", f[1][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;
const int N = 1e5 + 5;
int n, f[2][N];
long long a[N], mx, v1[N], v2[N];
void upd(int& x, int y) { x = (x < y ? x : y); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]), mx = max(mx, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = mx - a[i];
  memset(f[1], 0x3f, sizeof(int) * (n + 1));
  f[1][0] = 0;
  for (int i = 0; i < 64; ++i) {
    memset(f[i & 1], 0x3f, sizeof(int) * (n + 1));
    int c[2] = {0}, t = 0;
    for (int j = 1; j <= n; ++j) ++c[a[j] >> i & 1];
    upd(f[i & 1][0], f[~i & 1][0] + c[1]);
    upd(f[i & 1][c[1]], f[~i & 1][0] + c[0]);
    for (int j = 1; j <= n; ++j) {
      if (a[j] >> i & 1)
        --c[1], ++c[0], ++t;
      else
        --c[0], ++c[1];
      upd(f[i & 1][t], f[~i & 1][j] + c[1]);
      upd(f[i & 1][t + c[1]], f[~i & 1][j] + c[0]);
    }
    int id1 = 0, id2 = 0, id = 0;
    for (int j = 1; j <= n; ++j)
      (a[j] >> i & 1) ? v1[++id1] = a[j] : v2[++id2] = a[j];
    for (int j = 1; j <= id1; ++j) a[++id] = v1[j];
    for (int j = 1; j <= id2; ++j) a[++id] = v2[j];
  }
  printf("%d", f[1][0]);
}

```