#include <bits/stdc++.h>
using namespace std;
long long n, mx, a[200010], g[200100], g1[200010], f[63][100010], s1[100010],
    s0[100010], cnt, sz0, sz1;
void px(long long x) {
  cnt = 0;
  int i;
  for (i = 1; i <= n; i++)
    if ((a[g[i]] & (1ll << x)) == 0) g1[++cnt] = g[i];
  for (i = 1; i <= n; i++)
    if (a[g[i]] & (1ll << x)) g1[++cnt] = g[i];
  for (i = 1; i <= n; i++) g[i] = g1[i];
}
void ycl(long long x) {
  long long i;
  for (i = 1; i <= n; i++) {
    s0[i] = s0[i - 1];
    s1[i] = s1[i - 1];
    if (a[g[i]] & (1ll << x))
      s1[i]++;
    else
      s0[i]++;
  }
  sz0 = s0[n];
  sz1 = s1[n];
}
int main() {
  long long i, j, k, val, nt;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (i = 1; i <= n; i++) a[i] = mx - a[i];
  memset(f, 44, sizeof(f));
  for (i = 1; i <= n; i++) g[i] = i;
  f[0][0] = 0;
  for (i = 0; i < 62; i++) {
    if (i > 0) px(i - 1);
    ycl(i);
    for (j = 0; j <= n; j++)
      for (k = 0; k < 2; k++) {
        if (k == 0) {
          val = s1[n - j] + sz0 - s0[n - j];
          nt = sz1 - s1[n - j];
          f[i + 1][nt] = min(f[i + 1][nt], f[i][j] + val);
        } else {
          val = s0[n - j] + sz1 - s1[n - j];
          nt = n - s0[n - j];
          f[i + 1][nt] = min(f[i + 1][nt], f[i][j] + val);
        }
      }
  }
  printf("%lld", f[62][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;
long long n, mx, a[200010], g[200100], g1[200010], f[63][100010], s1[100010],
    s0[100010], cnt, sz0, sz1;
void px(long long x) {
  cnt = 0;
  int i;
  for (i = 1; i <= n; i++)
    if ((a[g[i]] & (1ll << x)) == 0) g1[++cnt] = g[i];
  for (i = 1; i <= n; i++)
    if (a[g[i]] & (1ll << x)) g1[++cnt] = g[i];
  for (i = 1; i <= n; i++) g[i] = g1[i];
}
void ycl(long long x) {
  long long i;
  for (i = 1; i <= n; i++) {
    s0[i] = s0[i - 1];
    s1[i] = s1[i - 1];
    if (a[g[i]] & (1ll << x))
      s1[i]++;
    else
      s0[i]++;
  }
  sz0 = s0[n];
  sz1 = s1[n];
}
int main() {
  long long i, j, k, val, nt;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (i = 1; i <= n; i++) a[i] = mx - a[i];
  memset(f, 44, sizeof(f));
  for (i = 1; i <= n; i++) g[i] = i;
  f[0][0] = 0;
  for (i = 0; i < 62; i++) {
    if (i > 0) px(i - 1);
    ycl(i);
    for (j = 0; j <= n; j++)
      for (k = 0; k < 2; k++) {
        if (k == 0) {
          val = s1[n - j] + sz0 - s0[n - j];
          nt = sz1 - s1[n - j];
          f[i + 1][nt] = min(f[i + 1][nt], f[i][j] + val);
        } else {
          val = s0[n - j] + sz1 - s1[n - j];
          nt = n - s0[n - j];
          f[i + 1][nt] = min(f[i + 1][nt], f[i][j] + val);
        }
      }
  }
  printf("%lld", f[62][0]);
}

```