#include <bits/stdc++.h>
using namespace std;
const long long N = 100005;
long long n, now, s[N], a[N], f[62][N], b[N];
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) a[i] = a[n] - a[i];
  memset(f, 0x3f3f3f3f, sizeof f);
  f[0][0] = 0;
  for (long long i = 0; i <= 60; i++) {
    for (long long j = 1; j <= n; j++)
      s[j] = s[j - 1] + ((a[j] & (1ll << i)) ? 1 : 0);
    for (long long j = 0; j <= n; j++) {
      long long v = j - s[j] + s[n] - s[j], cnt = s[j];
      f[i + 1][cnt] = min(f[i + 1][cnt], f[i][j] + v);
      v = s[j] + (n - j) - (s[n] - s[j]);
      cnt = j + s[n] - s[j];
      f[i + 1][cnt] = min(f[i + 1][cnt], f[i][j] + v);
    }
    long long k = 0;
    for (long long j = 1; j <= n; j++)
      if (a[j] & (1ll << i)) b[++k] = a[j];
    for (long long j = 1; j <= n; j++)
      if (!(a[j] & (1ll << i))) b[++k] = a[j];
    for (long long j = 1; j <= n; j++) a[j] = b[j];
  }
  printf("%lld\n", f[61][0]);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 long long N = 100005;
long long n, now, s[N], a[N], f[62][N], b[N];
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) a[i] = a[n] - a[i];
  memset(f, 0x3f3f3f3f, sizeof f);
  f[0][0] = 0;
  for (long long i = 0; i <= 60; i++) {
    for (long long j = 1; j <= n; j++)
      s[j] = s[j - 1] + ((a[j] & (1ll << i)) ? 1 : 0);
    for (long long j = 0; j <= n; j++) {
      long long v = j - s[j] + s[n] - s[j], cnt = s[j];
      f[i + 1][cnt] = min(f[i + 1][cnt], f[i][j] + v);
      v = s[j] + (n - j) - (s[n] - s[j]);
      cnt = j + s[n] - s[j];
      f[i + 1][cnt] = min(f[i + 1][cnt], f[i][j] + v);
    }
    long long k = 0;
    for (long long j = 1; j <= n; j++)
      if (a[j] & (1ll << i)) b[++k] = a[j];
    for (long long j = 1; j <= n; j++)
      if (!(a[j] & (1ll << i))) b[++k] = a[j];
    for (long long j = 1; j <= n; j++) a[j] = b[j];
  }
  printf("%lld\n", f[61][0]);
}

```