#include <bits/stdc++.h>
const int MAXN = 1e5 + 1;
int N;
long long a[MAXN], K;
int f[63][MAXN], id[MAXN];
int sum[2][MAXN];
bool cmp(int x, int y) { return (a[x] & K) < (a[y] & K); }
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cin >> N;
  for (int i = 1; i <= N; i++) {
    std::cin >> a[i];
  }
  std::sort(a + 1, a + N + 1);
  for (int i = 1; i <= N; i++) {
    a[i] = a[N] - a[i];
    id[i] = i;
  }
  memset(f, 63, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k < 62; ++k) {
    K = 1ll << k;
    --K;
    std::sort(id + 1, id + N + 1, cmp);
    for (int i = 1; i <= N; i++) {
      sum[0][i] = sum[0][i - 1];
      sum[1][i] = sum[1][i - 1];
      sum[a[id[i]] >> k & 1][i]++;
    }
    for (int i = 0; i <= N; i++) {
      int w, rever;
      w = sum[1][N - i] + sum[0][N] - sum[0][N - i];
      rever = sum[1][N] - sum[1][N - i];
      f[k + 1][rever] = std::min(f[k + 1][rever], f[k][i] + w);
      w = sum[0][N - i] + sum[1][N] - sum[1][N - i];
      rever = N - sum[0][N - i];
      f[k + 1][rever] = std::min(f[k + 1][rever], f[k][i] + w);
    }
  }
  std::cout << f[62][0] << '\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>
const int MAXN = 1e5 + 1;
int N;
long long a[MAXN], K;
int f[63][MAXN], id[MAXN];
int sum[2][MAXN];
bool cmp(int x, int y) { return (a[x] & K) < (a[y] & K); }
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cin >> N;
  for (int i = 1; i <= N; i++) {
    std::cin >> a[i];
  }
  std::sort(a + 1, a + N + 1);
  for (int i = 1; i <= N; i++) {
    a[i] = a[N] - a[i];
    id[i] = i;
  }
  memset(f, 63, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k < 62; ++k) {
    K = 1ll << k;
    --K;
    std::sort(id + 1, id + N + 1, cmp);
    for (int i = 1; i <= N; i++) {
      sum[0][i] = sum[0][i - 1];
      sum[1][i] = sum[1][i - 1];
      sum[a[id[i]] >> k & 1][i]++;
    }
    for (int i = 0; i <= N; i++) {
      int w, rever;
      w = sum[1][N - i] + sum[0][N] - sum[0][N - i];
      rever = sum[1][N] - sum[1][N - i];
      f[k + 1][rever] = std::min(f[k + 1][rever], f[k][i] + w);
      w = sum[0][N - i] + sum[1][N] - sum[1][N - i];
      rever = N - sum[0][N - i];
      f[k + 1][rever] = std::min(f[k + 1][rever], f[k][i] + w);
    }
  }
  std::cout << f[62][0] << '\n';
  return 0;
}

```