#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
long long dp[60][N], a[N], ans, mx;
int n, F[2][N], *f = F[0], *g = F[1], sz[2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i], mx = max(mx, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = mx - a[i], f[i] = i;
  memset(dp, 0x3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 0; i < 57; ++i) {
    sz[0] = sz[1];
    for (int j = 1; j <= n; ++j) ++sz[a[j] >> i & 1];
    int u = sz[1], v = 0;
    for (int j = 0; j <= n; ++j) {
      dp[i + 1][v] = min(dp[i + 1][v], dp[i][j] + u);
      dp[i + 1][u + v] = min(dp[i + 1][u + v], dp[i][j] + n - u);
      if (a[f[j + 1]] >> i & 1)
        ++v, --u;
      else
        ++u;
    }
    sz[0] += sz[1];
    for (int j = n; j; --j) g[sz[a[f[j]] >> i & 1]--] = f[j];
    swap(f, g);
    ans = dp[i + 1][0];
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in cpp for 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 = 100100;
long long dp[60][N], a[N], ans, mx;
int n, F[2][N], *f = F[0], *g = F[1], sz[2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i], mx = max(mx, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = mx - a[i], f[i] = i;
  memset(dp, 0x3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 0; i < 57; ++i) {
    sz[0] = sz[1];
    for (int j = 1; j <= n; ++j) ++sz[a[j] >> i & 1];
    int u = sz[1], v = 0;
    for (int j = 0; j <= n; ++j) {
      dp[i + 1][v] = min(dp[i + 1][v], dp[i][j] + u);
      dp[i + 1][u + v] = min(dp[i + 1][u + v], dp[i][j] + n - u);
      if (a[f[j + 1]] >> i & 1)
        ++v, --u;
      else
        ++u;
    }
    sz[0] += sz[1];
    for (int j = n; j; --j) g[sz[a[f[j]] >> i & 1]--] = f[j];
    swap(f, g);
    ans = dp[i + 1][0];
  }
  cout << ans << '\n';
  return 0;
}

```