#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n;
long long a[N], b[N];
int dp[2][N];
void Sort(long long bit) {
  int n1 = 0;
  for (int i = 0; i < n; i++) b[i] = a[i];
  for (int i = 0; i < n; i++) {
    if (b[i] & bit) a[n1++] = b[i];
  }
  for (int i = 0; i < n; i++) {
    if (!(b[i] & bit)) a[n1++] = b[i];
  }
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  memset(dp[0], INF, sizeof(dp[0]));
  dp[0][0] = 0;
  for (int i = 0; i < 62; i++) {
    bool f = i & 1;
    int cntl = 0, cntr = 0;
    memset(dp[f ^ 1], INF, sizeof(dp[f ^ 1]));
    for (int j = 0; j < n; j++) {
      if (a[j] & (1LL << i)) cntr++;
    }
    for (int j = 0; j <= n; j++) {
      if (dp[f][j] != INF) {
        dp[f ^ 1][j + cntr] =
            min(dp[f ^ 1][j + cntr], dp[f][j] + j - cntl + cntr);
        dp[f ^ 1][cntl] =
            min(dp[f ^ 1][cntl], dp[f][j] + cntl + (n - j - cntr));
      }
      if (a[j] & (1LL << i)) {
        cntl++;
        cntr--;
      }
    }
    if (f && (1LL << i) > mx) break;
    Sort(1LL << i);
  }
  printf("%d\n", dp[0][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 N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n;
long long a[N], b[N];
int dp[2][N];
void Sort(long long bit) {
  int n1 = 0;
  for (int i = 0; i < n; i++) b[i] = a[i];
  for (int i = 0; i < n; i++) {
    if (b[i] & bit) a[n1++] = b[i];
  }
  for (int i = 0; i < n; i++) {
    if (!(b[i] & bit)) a[n1++] = b[i];
  }
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  memset(dp[0], INF, sizeof(dp[0]));
  dp[0][0] = 0;
  for (int i = 0; i < 62; i++) {
    bool f = i & 1;
    int cntl = 0, cntr = 0;
    memset(dp[f ^ 1], INF, sizeof(dp[f ^ 1]));
    for (int j = 0; j < n; j++) {
      if (a[j] & (1LL << i)) cntr++;
    }
    for (int j = 0; j <= n; j++) {
      if (dp[f][j] != INF) {
        dp[f ^ 1][j + cntr] =
            min(dp[f ^ 1][j + cntr], dp[f][j] + j - cntl + cntr);
        dp[f ^ 1][cntl] =
            min(dp[f ^ 1][cntl], dp[f][j] + cntl + (n - j - cntr));
      }
      if (a[j] & (1LL << i)) {
        cntl++;
        cntr--;
      }
    }
    if (f && (1LL << i) > mx) break;
    Sort(1LL << i);
  }
  printf("%d\n", dp[0][0]);
  return 0;
}

```