#include <bits/stdc++.h>
using namespace std;
long long a[100005];
long long dp[65][100005];
long long k = 0;
bool cmp(long long a, long long b) {
  long long t = (1ll << k) - 1;
  return (a & t) > (b & t);
}
int main() {
  int n;
  scanf("%d", &n);
  long long mx = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = mx - a[i];
  }
  memset(dp, 0x3f3f3f3f, sizeof(dp));
  for (k = 0; k < 63; k++) {
    sort(a + 1, a + n + 1, cmp);
    int OneNum = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] >> k & 1) OneNum++;
    }
    int one = 0, zero = 0;
    if (k != 0) {
      for (int i = 1; i <= n; i++) {
        if (a[i] >> k & 1)
          one++;
        else
          zero++;
        dp[k][one] = min(dp[k][one], dp[k - 1][i] + OneNum - one + zero);
        dp[k][i + OneNum - one] = min(dp[k][i + OneNum - one],
                                      dp[k - 1][i] + one + n - OneNum - zero);
      }
      dp[k][0] = min(dp[k][0], dp[k - 1][0] + OneNum);
      dp[k][OneNum] = min(dp[k][OneNum], dp[k - 1][0] + n - OneNum);
    } else {
      dp[k][0] = min(dp[k][0], 1LL * OneNum);
      dp[k][OneNum] = min(dp[k][OneNum], 1LL * (n - OneNum));
    }
  }
  printf("%d\n", dp[62][0]);
  return 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;
long long a[100005];
long long dp[65][100005];
long long k = 0;
bool cmp(long long a, long long b) {
  long long t = (1ll << k) - 1;
  return (a & t) > (b & t);
}
int main() {
  int n;
  scanf("%d", &n);
  long long mx = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = mx - a[i];
  }
  memset(dp, 0x3f3f3f3f, sizeof(dp));
  for (k = 0; k < 63; k++) {
    sort(a + 1, a + n + 1, cmp);
    int OneNum = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] >> k & 1) OneNum++;
    }
    int one = 0, zero = 0;
    if (k != 0) {
      for (int i = 1; i <= n; i++) {
        if (a[i] >> k & 1)
          one++;
        else
          zero++;
        dp[k][one] = min(dp[k][one], dp[k - 1][i] + OneNum - one + zero);
        dp[k][i + OneNum - one] = min(dp[k][i + OneNum - one],
                                      dp[k - 1][i] + one + n - OneNum - zero);
      }
      dp[k][0] = min(dp[k][0], dp[k - 1][0] + OneNum);
      dp[k][OneNum] = min(dp[k][OneNum], dp[k - 1][0] + n - OneNum);
    } else {
      dp[k][0] = min(dp[k][0], 1LL * OneNum);
      dp[k][OneNum] = min(dp[k][OneNum], 1LL * (n - OneNum));
    }
  }
  printf("%d\n", dp[62][0]);
  return 0;
}

```