#include <bits/stdc++.h>
using namespace std;
const int maxn = 100020;
const int maxbit = 60;
const long long inf = (long long)(1e17 + 1);
long long a[maxn], dp[maxbit + 2][maxn];
int zero, one, zeroc, onec, n, tmp;
bool cmp(long long i, long long j) {
  long long x = i & ((1LL << tmp) - 1);
  long long y = j & ((1LL << tmp) - 1);
  return x > y;
}
void Initialization() {
  int odd = 0;
  for (int i = 1; i <= n; i++) {
    if ((a[i] & 1) != 0) odd++;
  }
  dp[0][0] = n - odd;
  if (odd == 0)
    dp[0][odd] = 0;
  else
    dp[0][odd] = odd;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i <= maxbit; i++) {
    for (int j = 0; j <= n; j++) dp[i][j] = inf;
  }
  Initialization();
  for (int i = 1; i <= maxbit; i++) {
    tmp = i;
    sort(a + 1, a + n + 1, cmp);
    zero = 0, one = 0, zeroc = 0, onec = 0;
    for (int j = 1; j <= n; j++) {
      if ((a[j] >> i) & 1)
        one++;
      else
        zero++;
    }
    for (int j = 0; j <= n; j++) {
      if (j) {
        if ((a[j] >> i) & 1) {
          one--;
          onec++;
        } else {
          zero--;
          zeroc++;
        }
      }
      dp[i][onec] = min(dp[i][onec], dp[i - 1][j] + zero + onec);
      dp[i][n - zero] = min(dp[i][n - zero], dp[i - 1][j] + one + zeroc);
    }
  }
  cout << min(dp[maxbit][n], dp[maxbit][0]);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 maxn = 100020;
const int maxbit = 60;
const long long inf = (long long)(1e17 + 1);
long long a[maxn], dp[maxbit + 2][maxn];
int zero, one, zeroc, onec, n, tmp;
bool cmp(long long i, long long j) {
  long long x = i & ((1LL << tmp) - 1);
  long long y = j & ((1LL << tmp) - 1);
  return x > y;
}
void Initialization() {
  int odd = 0;
  for (int i = 1; i <= n; i++) {
    if ((a[i] & 1) != 0) odd++;
  }
  dp[0][0] = n - odd;
  if (odd == 0)
    dp[0][odd] = 0;
  else
    dp[0][odd] = odd;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i <= maxbit; i++) {
    for (int j = 0; j <= n; j++) dp[i][j] = inf;
  }
  Initialization();
  for (int i = 1; i <= maxbit; i++) {
    tmp = i;
    sort(a + 1, a + n + 1, cmp);
    zero = 0, one = 0, zeroc = 0, onec = 0;
    for (int j = 1; j <= n; j++) {
      if ((a[j] >> i) & 1)
        one++;
      else
        zero++;
    }
    for (int j = 0; j <= n; j++) {
      if (j) {
        if ((a[j] >> i) & 1) {
          one--;
          onec++;
        } else {
          zero--;
          zeroc++;
        }
      }
      dp[i][onec] = min(dp[i][onec], dp[i - 1][j] + zero + onec);
      dp[i][n - zero] = min(dp[i][n - zero], dp[i - 1][j] + one + zeroc);
    }
  }
  cout << min(dp[maxbit][n], dp[maxbit][0]);
  return 0;
}

```