#include <bits/stdc++.h>
using namespace std;
const int oo = 27081993;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  long long a[100100];
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long maxA = *max_element(a, a + n);
  for (int i = 0; i < n; i++) a[i] = maxA - a[i];
  vector<int> f(n + 1, oo);
  f[0] = 0;
  for (int bit = 0; bit < 60; bit++) {
    vector<long long> b;
    vector<int> cnt(3, 0);
    for (int i = 0; i < n; i++) {
      b.push_back(a[i] % (1LL << bit + 1));
      cnt[a[i] >> bit & 1]++;
    }
    sort(b.begin(), b.end());
    auto compare = [&](long long x, long long y) {
      return x % (1LL << bit) < y % (1LL << bit);
    };
    vector<int> oldF(f);
    f.assign(n + 1, oo);
    int oldCnt0 = cnt[0];
    for (int i = 0; i <= n; i++) {
      if (i && bit) {
        if (!cnt[0] ||
            (cnt[1] && compare(b[cnt[0] - 1], b[oldCnt0 + cnt[1] - 1]))) {
          cnt[1]--;
          cnt[2]++;
        } else
          cnt[0]--;
      }
      f[cnt[2]] = min(f[cnt[2]], oldF[i] + n - cnt[0] - cnt[2]);
      f[n - cnt[0]] = min(f[n - cnt[0]], oldF[i] + cnt[0] + cnt[2]);
    }
  }
  cout << f[0] << endl;
}

### Prompt

In cpp, your task is to solve 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 oo = 27081993;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  long long a[100100];
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long maxA = *max_element(a, a + n);
  for (int i = 0; i < n; i++) a[i] = maxA - a[i];
  vector<int> f(n + 1, oo);
  f[0] = 0;
  for (int bit = 0; bit < 60; bit++) {
    vector<long long> b;
    vector<int> cnt(3, 0);
    for (int i = 0; i < n; i++) {
      b.push_back(a[i] % (1LL << bit + 1));
      cnt[a[i] >> bit & 1]++;
    }
    sort(b.begin(), b.end());
    auto compare = [&](long long x, long long y) {
      return x % (1LL << bit) < y % (1LL << bit);
    };
    vector<int> oldF(f);
    f.assign(n + 1, oo);
    int oldCnt0 = cnt[0];
    for (int i = 0; i <= n; i++) {
      if (i && bit) {
        if (!cnt[0] ||
            (cnt[1] && compare(b[cnt[0] - 1], b[oldCnt0 + cnt[1] - 1]))) {
          cnt[1]--;
          cnt[2]++;
        } else
          cnt[0]--;
      }
      f[cnt[2]] = min(f[cnt[2]], oldF[i] + n - cnt[0] - cnt[2]);
      f[n - cnt[0]] = min(f[n - cnt[0]], oldF[i] + cnt[0] + cnt[2]);
    }
  }
  cout << f[0] << endl;
}

```