#include <bits/stdc++.h>
using namespace std;
const int MxN = (int)1e5 + 9;
const int MxI = 60;
long long a[MxN + 9];
int dp[2][MxN + 9];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; ++i) a[i] = a[n - 1] - a[i];
  for (int i = MxI; i >= 0; --i) {
    sort(a, a + n, [&i](const long long x, const long long y) {
      const long long mask = (1LL << i) - 1;
      return (x & mask) < (y & mask);
    });
    array<long long, 3> cnt;
    cnt.fill(0);
    for (int j = 0; j < n; ++j) ++cnt[a[j] & (1LL << i) ? 1 : 0];
    for (int j = 0; j <= n; ++j) {
      int &res = dp[i & 1][j];
      if (i == MxI)
        res = j;
      else {
        const int cost1 = cnt[1];
        const int cost2 = cnt[0] + cnt[2];
        const int res1 = cost1 + dp[~i & 1][cnt[2]];
        const int res2 = cost2 + dp[~i & 1][cnt[1] + cnt[2]];
        res = min(res1, res2);
      }
      if (j < n) {
        --cnt[a[n - 1 - j] & (1LL << i) ? 1 : 0];
        ++cnt[a[n - 1 - j] & (1LL << i) ? 2 : 1];
        for (int x : cnt) assert(x >= 0);
      }
    }
  }
  cout << dp[0][0];
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
const int MxN = (int)1e5 + 9;
const int MxI = 60;
long long a[MxN + 9];
int dp[2][MxN + 9];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; ++i) a[i] = a[n - 1] - a[i];
  for (int i = MxI; i >= 0; --i) {
    sort(a, a + n, [&i](const long long x, const long long y) {
      const long long mask = (1LL << i) - 1;
      return (x & mask) < (y & mask);
    });
    array<long long, 3> cnt;
    cnt.fill(0);
    for (int j = 0; j < n; ++j) ++cnt[a[j] & (1LL << i) ? 1 : 0];
    for (int j = 0; j <= n; ++j) {
      int &res = dp[i & 1][j];
      if (i == MxI)
        res = j;
      else {
        const int cost1 = cnt[1];
        const int cost2 = cnt[0] + cnt[2];
        const int res1 = cost1 + dp[~i & 1][cnt[2]];
        const int res2 = cost2 + dp[~i & 1][cnt[1] + cnt[2]];
        res = min(res1, res2);
      }
      if (j < n) {
        --cnt[a[n - 1 - j] & (1LL << i) ? 1 : 0];
        ++cnt[a[n - 1 - j] & (1LL << i) ? 2 : 1];
        for (int x : cnt) assert(x >= 0);
      }
    }
  }
  cout << dp[0][0];
  return 0;
}

```