#include <bits/stdc++.h>
using namespace std;
const int N = 60;
long long sl, fh, a[100010], b[N][100010];
int n, dp[N][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 count(int i, long long x, int p) {
  long long S = 1ll << (i - 1);
  if (x < S)
    return upper_bound(b[i], b[i] + n, x + S) - b[i] - p;
  else
    return upper_bound(b[i], b[i] + n, x - S) - b[i] + n - p;
}
int main() {
  n = rd();
  long long x;
  for (int i = 1; i <= n; ++i) a[i] = rd();
  for (int i = 1; i < N; ++i) {
    x = (1ll << i) - 1;
    for (int j = 1; j <= n; ++j) b[i][j - 1] = a[j] & x;
    sort(b[i], b[i] + n);
  }
  memset(dp, 127 / 3, sizeof(dp));
  dp[0][n] = 0;
  int p;
  for (int i = 1; i < N; ++i)
    for (int j = 0; j <= n; ++j)
      if (dp[i - 1][j] <= 1e8) {
        for (long long v = 0; v < 2; ++v) {
          x = v << (i - 1);
          if (j) x += b[i - 1][j - 1];
          p = upper_bound(b[i], b[i] + n, x) - b[i];
          dp[i][p] = min(dp[i][p], dp[i - 1][j] + count(i, x, p));
        }
      }
  printf("%d\n", dp[N - 1][n]);
  return 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 int N = 60;
long long sl, fh, a[100010], b[N][100010];
int n, dp[N][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 count(int i, long long x, int p) {
  long long S = 1ll << (i - 1);
  if (x < S)
    return upper_bound(b[i], b[i] + n, x + S) - b[i] - p;
  else
    return upper_bound(b[i], b[i] + n, x - S) - b[i] + n - p;
}
int main() {
  n = rd();
  long long x;
  for (int i = 1; i <= n; ++i) a[i] = rd();
  for (int i = 1; i < N; ++i) {
    x = (1ll << i) - 1;
    for (int j = 1; j <= n; ++j) b[i][j - 1] = a[j] & x;
    sort(b[i], b[i] + n);
  }
  memset(dp, 127 / 3, sizeof(dp));
  dp[0][n] = 0;
  int p;
  for (int i = 1; i < N; ++i)
    for (int j = 0; j <= n; ++j)
      if (dp[i - 1][j] <= 1e8) {
        for (long long v = 0; v < 2; ++v) {
          x = v << (i - 1);
          if (j) x += b[i - 1][j - 1];
          p = upper_bound(b[i], b[i] + n, x) - b[i];
          dp[i][p] = min(dp[i][p], dp[i - 1][j] + count(i, x, p));
        }
      }
  printf("%d\n", dp[N - 1][n]);
  return 0;
}

```