#include <bits/stdc++.h>
using namespace std;
const long long oo = 1LL << 62;
int n;
long long a[100005], dp[70][100005], mx;
int main() {
  ios_base::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];
  for (int i = 0; i <= 64; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = oo;
  dp[0][0] = 0;
  vector<int> f;
  for (int i = 1; i <= n; i++) f.push_back(i);
  for (int bit = 0; bit <= 62; bit++) {
    if (bit > 0) {
      vector<int> nf;
      for (int v : f)
        if (!(a[v] & (1LL << (bit - 1)))) nf.push_back(v);
      for (int v : f)
        if ((a[v] & (1LL << (bit - 1)))) nf.push_back(v);
      f = nf;
    }
    vector<int> prefZeroes(n + 1);
    vector<int> prefOnes(n + 1);
    prefZeroes[0] = prefOnes[0] = 0;
    for (int i = 1; i <= n; i++) {
      prefZeroes[i] = prefZeroes[i - 1];
      prefOnes[i] = prefOnes[i - 1];
      if (a[f[i - 1]] & (1LL << bit))
        prefOnes[i]++;
      else
        prefZeroes[i]++;
    }
    int totalZeroes = prefZeroes[n];
    int totalOnes = prefOnes[n];
    for (int sufTake = 0; sufTake <= n; sufTake++)
      if (dp[bit][sufTake] != oo) {
        int numOnes =
            totalZeroes - prefZeroes[n - sufTake] + prefOnes[n - sufTake];
        int carryOnes = totalOnes - prefOnes[n - sufTake];
        dp[bit + 1][carryOnes] =
            min(dp[bit + 1][carryOnes], dp[bit][sufTake] + numOnes);
        numOnes = totalOnes - prefOnes[n - sufTake] + prefZeroes[n - sufTake];
        carryOnes = n - prefZeroes[n - sufTake];
        dp[bit + 1][carryOnes] =
            min(dp[bit + 1][carryOnes], dp[bit][sufTake] + numOnes);
      }
  }
  cout << dp[63][0];
}

### Prompt

Generate 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>
using namespace std;
const long long oo = 1LL << 62;
int n;
long long a[100005], dp[70][100005], mx;
int main() {
  ios_base::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];
  for (int i = 0; i <= 64; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = oo;
  dp[0][0] = 0;
  vector<int> f;
  for (int i = 1; i <= n; i++) f.push_back(i);
  for (int bit = 0; bit <= 62; bit++) {
    if (bit > 0) {
      vector<int> nf;
      for (int v : f)
        if (!(a[v] & (1LL << (bit - 1)))) nf.push_back(v);
      for (int v : f)
        if ((a[v] & (1LL << (bit - 1)))) nf.push_back(v);
      f = nf;
    }
    vector<int> prefZeroes(n + 1);
    vector<int> prefOnes(n + 1);
    prefZeroes[0] = prefOnes[0] = 0;
    for (int i = 1; i <= n; i++) {
      prefZeroes[i] = prefZeroes[i - 1];
      prefOnes[i] = prefOnes[i - 1];
      if (a[f[i - 1]] & (1LL << bit))
        prefOnes[i]++;
      else
        prefZeroes[i]++;
    }
    int totalZeroes = prefZeroes[n];
    int totalOnes = prefOnes[n];
    for (int sufTake = 0; sufTake <= n; sufTake++)
      if (dp[bit][sufTake] != oo) {
        int numOnes =
            totalZeroes - prefZeroes[n - sufTake] + prefOnes[n - sufTake];
        int carryOnes = totalOnes - prefOnes[n - sufTake];
        dp[bit + 1][carryOnes] =
            min(dp[bit + 1][carryOnes], dp[bit][sufTake] + numOnes);
        numOnes = totalOnes - prefOnes[n - sufTake] + prefZeroes[n - sufTake];
        carryOnes = n - prefZeroes[n - sufTake];
        dp[bit + 1][carryOnes] =
            min(dp[bit + 1][carryOnes], dp[bit][sufTake] + numOnes);
      }
  }
  cout << dp[63][0];
}

```