#include <bits/stdc++.h>
using namespace std;
long long now, f[2][100005], cntr, cntl, n, lim, a[100005], b[100005];
long long gets() {
  long long v = 0;
  char ch;
  bool f = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = 1;
  v = ch - 48;
  while (isdigit(ch = getchar())) v = (v << 1) + (v << 3) + ch - 48;
  return f ? -v : v;
}
void Sort(long long bit) {
  long long tot = 0;
  for (long long i = 1; i <= n; i++) b[i] = a[i];
  for (long long i = 1; i <= n; i++)
    if (b[i] & bit) a[++tot] = b[i];
  for (long long i = 1; i <= n; i++)
    if (!(b[i] & bit)) a[++tot] = b[i];
}
void test(long long bit) {
  for (long long i = 0; i <= n; i++) f[now ^ 1][i] = 1000000000;
  cntl = cntr = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] & (1ll << bit)) cntr++;
  }
  for (long long i = 0; i <= n; i++) {
    if (a[i] & (1ll << bit)) cntl++, cntr--;
    if (f[now][i] != 1000000000) {
      f[now ^ 1][i + cntr] =
          min(f[now ^ 1][i + cntr], f[now][i] + i + cntr - cntl);
      f[now ^ 1][cntl] = min(f[now ^ 1][cntl], f[now][i] + cntl + n - cntr - i);
    }
  }
  now ^= 1;
  Sort(1ll << bit);
}
int main() {
  n = gets();
  lim = 60;
  now = 1;
  for (long long i = 1; i <= n; i++) a[i] = gets();
  for (long long i = 1; i <= n; i++) f[1][i] = 1000000000;
  f[1][0] = 0;
  for (long long i = 0; i <= lim; i++) test(i);
  printf("%d\n", f[now][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;
long long now, f[2][100005], cntr, cntl, n, lim, a[100005], b[100005];
long long gets() {
  long long v = 0;
  char ch;
  bool f = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = 1;
  v = ch - 48;
  while (isdigit(ch = getchar())) v = (v << 1) + (v << 3) + ch - 48;
  return f ? -v : v;
}
void Sort(long long bit) {
  long long tot = 0;
  for (long long i = 1; i <= n; i++) b[i] = a[i];
  for (long long i = 1; i <= n; i++)
    if (b[i] & bit) a[++tot] = b[i];
  for (long long i = 1; i <= n; i++)
    if (!(b[i] & bit)) a[++tot] = b[i];
}
void test(long long bit) {
  for (long long i = 0; i <= n; i++) f[now ^ 1][i] = 1000000000;
  cntl = cntr = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] & (1ll << bit)) cntr++;
  }
  for (long long i = 0; i <= n; i++) {
    if (a[i] & (1ll << bit)) cntl++, cntr--;
    if (f[now][i] != 1000000000) {
      f[now ^ 1][i + cntr] =
          min(f[now ^ 1][i + cntr], f[now][i] + i + cntr - cntl);
      f[now ^ 1][cntl] = min(f[now ^ 1][cntl], f[now][i] + cntl + n - cntr - i);
    }
  }
  now ^= 1;
  Sort(1ll << bit);
}
int main() {
  n = gets();
  lim = 60;
  now = 1;
  for (long long i = 1; i <= n; i++) a[i] = gets();
  for (long long i = 1; i <= n; i++) f[1][i] = 1000000000;
  f[1][0] = 0;
  for (long long i = 0; i <= lim; i++) test(i);
  printf("%d\n", f[now][0]);
  return 0;
}

```