#include <bits/stdc++.h>
using namespace std;
const int N = 100050, L = 60;
long long a[N], mx;
int dp[L][N], ord[N], tmp[N], cnt[2], n;
void ckmx(long long &x, long long y) { x = max(x, y); }
void ckmn(int &x, int y) { x = min(x, y); }
int main() {
  scanf("%i", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), ckmx(mx, a[i]);
  for (int i = 1; i <= n; i++) a[i] = mx - a[i], ord[i] = i;
  for (int i = 0; i < L; i++)
    for (int j = 0; j < N; j++) dp[i][j] = 1e9;
  dp[0][0] = 0;
  for (int j = 0; j < L - 1; j++) {
    for (int i = 1; i <= n; i++) cnt[a[i] >> j & 1]++;
    int o = cnt[1], l = 0;
    for (int i = 0; i <= n; i++) {
      ckmn(dp[j + 1][l], dp[j][i] + o);
      ckmn(dp[j + 1][l + o], dp[j][i] + n - o);
      if (i == n) break;
      if (a[ord[i + 1]] >> j & 1)
        o--, l++;
      else
        o++;
    }
    cnt[0] += cnt[1];
    for (int i = n; i >= 1; i--) tmp[cnt[a[ord[i]] >> j & 1]--] = ord[i];
    for (int i = 1; i <= n; i++) ord[i] = tmp[i];
    cnt[0] = cnt[1] = 0;
  }
  printf("%i\n", dp[L - 1][0]);
  return 0;
}

### Prompt

Develop a solution in Cpp 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 = 100050, L = 60;
long long a[N], mx;
int dp[L][N], ord[N], tmp[N], cnt[2], n;
void ckmx(long long &x, long long y) { x = max(x, y); }
void ckmn(int &x, int y) { x = min(x, y); }
int main() {
  scanf("%i", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), ckmx(mx, a[i]);
  for (int i = 1; i <= n; i++) a[i] = mx - a[i], ord[i] = i;
  for (int i = 0; i < L; i++)
    for (int j = 0; j < N; j++) dp[i][j] = 1e9;
  dp[0][0] = 0;
  for (int j = 0; j < L - 1; j++) {
    for (int i = 1; i <= n; i++) cnt[a[i] >> j & 1]++;
    int o = cnt[1], l = 0;
    for (int i = 0; i <= n; i++) {
      ckmn(dp[j + 1][l], dp[j][i] + o);
      ckmn(dp[j + 1][l + o], dp[j][i] + n - o);
      if (i == n) break;
      if (a[ord[i + 1]] >> j & 1)
        o--, l++;
      else
        o++;
    }
    cnt[0] += cnt[1];
    for (int i = n; i >= 1; i--) tmp[cnt[a[ord[i]] >> j & 1]--] = ord[i];
    for (int i = 1; i <= n; i++) ord[i] = tmp[i];
    cnt[0] = cnt[1] = 0;
  }
  printf("%i\n", dp[L - 1][0]);
  return 0;
}

```