#include <bits/stdc++.h>
using namespace std;
const int N = 60;
long long sl, fh, a[100010], s[100010], p[100010];
int n, dp[N + 5][100010];
long long rd() {
  sl = 0;
  fh = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') sl = sl * 10 + ch - '0', ch = getchar();
  return sl * fh;
}
int main() {
  n = rd();
  long long M = 0;
  for (int i = 1; i <= n; ++i) a[i] = rd(), M = max(M, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = M - a[i];
  memset(dp, 127 / 3, sizeof(dp));
  dp[0][n] = 0;
  int px, py, sx, sy;
  for (int i = 0; i < N; ++i) {
    for (int j = 1; j <= n; ++j) p[j] = p[j - 1] + ((a[j] >> i) & 1);
    for (int j = 0; j <= n; ++j)
      if (dp[i][j] <= 1e8) {
        px = p[j];
        py = j - px;
        sx = p[n] - px;
        sy = n - j - sx;
        dp[i + 1][n - sx] = min(dp[i + 1][n - sx], dp[i][j] + px + sy);
        dp[i + 1][py] = min(dp[i + 1][py], dp[i][j] + py + sx);
      }
    px = n - p[n];
    py = n;
    for (int j = n; j; --j)
      if ((a[j] >> i) & 1)
        s[py--] = a[j];
      else
        s[px--] = a[j];
    for (int j = 1; j <= n; ++j) a[j] = s[j];
  }
  printf("%d\n", dp[N][n]);
  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 = 60;
long long sl, fh, a[100010], s[100010], p[100010];
int n, dp[N + 5][100010];
long long rd() {
  sl = 0;
  fh = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') sl = sl * 10 + ch - '0', ch = getchar();
  return sl * fh;
}
int main() {
  n = rd();
  long long M = 0;
  for (int i = 1; i <= n; ++i) a[i] = rd(), M = max(M, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = M - a[i];
  memset(dp, 127 / 3, sizeof(dp));
  dp[0][n] = 0;
  int px, py, sx, sy;
  for (int i = 0; i < N; ++i) {
    for (int j = 1; j <= n; ++j) p[j] = p[j - 1] + ((a[j] >> i) & 1);
    for (int j = 0; j <= n; ++j)
      if (dp[i][j] <= 1e8) {
        px = p[j];
        py = j - px;
        sx = p[n] - px;
        sy = n - j - sx;
        dp[i + 1][n - sx] = min(dp[i + 1][n - sx], dp[i][j] + px + sy);
        dp[i + 1][py] = min(dp[i + 1][py], dp[i][j] + py + sx);
      }
    px = n - p[n];
    py = n;
    for (int j = n; j; --j)
      if ((a[j] >> i) & 1)
        s[py--] = a[j];
      else
        s[px--] = a[j];
    for (int j = 1; j <= n; ++j) a[j] = s[j];
  }
  printf("%d\n", dp[N][n]);
  return 0;
}

```