#include <bits/stdc++.h>
using namespace std;
int n, k, f[65][100005], cnt[2][100005];
long long a[100005];
bool cmp(long long x, long long y) {
  return (x & (1ll << k) - 1) < (y & (1ll << k) - 1);
}
void chkmin(int &a, int b) { a > b && (a = b); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  memset(f, 0x3f, sizeof f);
  f[0][0] = 0;
  for (k = 0; k <= 58; k++) {
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; i++) {
      cnt[0][i] = cnt[0][i - 1], cnt[1][i] = cnt[1][i - 1];
      cnt[a[i] >> k & 1][i]++;
    }
    for (int i = 0; i <= n; i++)
      if (f[k][i] < 0x3f3f3f3f) {
        chkmin(f[k + 1][cnt[1][n] - cnt[1][n - i]],
               f[k][i] + cnt[1][n - i] + cnt[0][n] - cnt[0][n - i]);
        chkmin(f[k + 1][n - cnt[0][n - i]],
               f[k][i] + cnt[0][n - i] + cnt[1][n] - cnt[1][n - i]);
      }
  }
  printf("%d\n", f[59][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;
int n, k, f[65][100005], cnt[2][100005];
long long a[100005];
bool cmp(long long x, long long y) {
  return (x & (1ll << k) - 1) < (y & (1ll << k) - 1);
}
void chkmin(int &a, int b) { a > b && (a = b); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  memset(f, 0x3f, sizeof f);
  f[0][0] = 0;
  for (k = 0; k <= 58; k++) {
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; i++) {
      cnt[0][i] = cnt[0][i - 1], cnt[1][i] = cnt[1][i - 1];
      cnt[a[i] >> k & 1][i]++;
    }
    for (int i = 0; i <= n; i++)
      if (f[k][i] < 0x3f3f3f3f) {
        chkmin(f[k + 1][cnt[1][n] - cnt[1][n - i]],
               f[k][i] + cnt[1][n - i] + cnt[0][n] - cnt[0][n - i]);
        chkmin(f[k + 1][n - cnt[0][n - i]],
               f[k][i] + cnt[0][n - i] + cnt[1][n] - cnt[1][n - i]);
      }
  }
  printf("%d\n", f[59][0]);
}

```