#include <bits/stdc++.h>
const int inf = 1e9;
long long a[100005], f[65][100005], U;
int n, s[100005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  long long M = *std::max_element(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) a[i] = M - a[i];
  for (int i = 0; i <= 63; ++i)
    for (int j = 0; j <= n; ++j) f[i][j] = inf;
  f[0][0] = 0;
  for (int i = 0; i <= 62; ++i) {
    for (int j = 1; j <= n; ++j) s[j] = s[j - 1] + ((a[n - j + 1] >> i) & 1);
    U = (1ll << (i + 1)) - 1;
    std::sort(a + 1, a + n + 1,
              [](long long a, long long b) { return (a & U) < (b & U); });
    int sum = s[n];
    for (int j = 0; j <= n; ++j) {
      int c11 = s[j];
      int c10 = sum - s[j];
      int c01 = j - s[j];
      int c00 = (n - j) - (sum - s[j]);
      f[i + 1][c11] = std::min(f[i + 1][c11], f[i][j] + c10 + c01);
      f[i + 1][c11 + c10 + c01] =
          std::min(f[i + 1][c11 + c10 + c01], f[i][j] + c11 + c00);
    }
  }
  printf("%lld", f[63][0]);
  return 0;
}

### Prompt

Your task is to create 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>
const int inf = 1e9;
long long a[100005], f[65][100005], U;
int n, s[100005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  long long M = *std::max_element(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) a[i] = M - a[i];
  for (int i = 0; i <= 63; ++i)
    for (int j = 0; j <= n; ++j) f[i][j] = inf;
  f[0][0] = 0;
  for (int i = 0; i <= 62; ++i) {
    for (int j = 1; j <= n; ++j) s[j] = s[j - 1] + ((a[n - j + 1] >> i) & 1);
    U = (1ll << (i + 1)) - 1;
    std::sort(a + 1, a + n + 1,
              [](long long a, long long b) { return (a & U) < (b & U); });
    int sum = s[n];
    for (int j = 0; j <= n; ++j) {
      int c11 = s[j];
      int c10 = sum - s[j];
      int c01 = j - s[j];
      int c00 = (n - j) - (sum - s[j]);
      f[i + 1][c11] = std::min(f[i + 1][c11], f[i][j] + c10 + c01);
      f[i + 1][c11 + c10 + c01] =
          std::min(f[i + 1][c11 + c10 + c01], f[i][j] + c11 + c00);
    }
  }
  printf("%lld", f[63][0]);
  return 0;
}

```