#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int inf = 1e9;
long long gi() {
  long long x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, f[63][N], s[N];
long long a[N];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) a[i] = gi();
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  memset(f, 0x3f, sizeof(f)), f[0][0] = 0;
  for (int i = 1; i <= 60; i++) {
    for (int j = 1; j <= n; j++) s[j] = s[j - 1] + (a[j] >> (i - 1) & 1);
    for (int j = 0; j <= n; j++) {
      if (f[i - 1][j] > inf) continue;
      int cnt = j - s[j] + s[n] - s[j], k = s[j];
      f[i][k] = min(f[i][k], f[i - 1][j] + cnt);
      cnt = n - cnt, k = j + s[n] - s[j];
      f[i][k] = min(f[i][k], f[i - 1][j] + cnt);
    }
    sort(a + 1, a + n + 1, [&](long long x, long long y) {
      return (x & ((1ll << i) - 1)) > (y & ((1ll << i) - 1));
    });
  }
  cout << f[60][0];
  return 0;
}

### Prompt

Please formulate 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 + 10;
const int inf = 1e9;
long long gi() {
  long long x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, f[63][N], s[N];
long long a[N];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) a[i] = gi();
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  memset(f, 0x3f, sizeof(f)), f[0][0] = 0;
  for (int i = 1; i <= 60; i++) {
    for (int j = 1; j <= n; j++) s[j] = s[j - 1] + (a[j] >> (i - 1) & 1);
    for (int j = 0; j <= n; j++) {
      if (f[i - 1][j] > inf) continue;
      int cnt = j - s[j] + s[n] - s[j], k = s[j];
      f[i][k] = min(f[i][k], f[i - 1][j] + cnt);
      cnt = n - cnt, k = j + s[n] - s[j];
      f[i][k] = min(f[i][k], f[i - 1][j] + cnt);
    }
    sort(a + 1, a + n + 1, [&](long long x, long long y) {
      return (x & ((1ll << i) - 1)) > (y & ((1ll << i) - 1));
    });
  }
  cout << f[60][0];
  return 0;
}

```