#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, dp[70][N];
long long a[N];
inline void umn(int &x, int y) { (x > y) && (x = y); }
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%lld", a + i);
  sort(a + 1, a + n + 1);
  for (int i = (1); i <= (n); ++i) a[i] = a[n] - a[i];
  for (int i = (0); i <= (61); ++i)
    for (int j = (0); j <= (n); ++j) dp[i][j] = 1e9;
  dp[0][0] = 0;
  for (int i = (0); i <= (60); ++i) {
    int cnt = 0, cur = 0;
    sort(a + 1, a + n + 1, [&](long long x, long long y) {
      return (x & ((1ll << i) - 1)) > (y & ((1ll << i) - 1));
    });
    for (int j = (1); j <= (n); ++j) cnt += a[j] >> i & 1;
    for (int j = (0); j <= (n); ++j) {
      cur += a[j] >> i & 1;
      int A = cur, B = j - A, C = cnt - cur, D = n - j - C;
      umn(dp[i + 1][A + B + C], dp[i][j] + A + D);
      umn(dp[i + 1][A], dp[i][j] + B + C);
    }
  }
  printf("%d\n", dp[61][0]);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int N = 1e5 + 10;
int n, dp[70][N];
long long a[N];
inline void umn(int &x, int y) { (x > y) && (x = y); }
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%lld", a + i);
  sort(a + 1, a + n + 1);
  for (int i = (1); i <= (n); ++i) a[i] = a[n] - a[i];
  for (int i = (0); i <= (61); ++i)
    for (int j = (0); j <= (n); ++j) dp[i][j] = 1e9;
  dp[0][0] = 0;
  for (int i = (0); i <= (60); ++i) {
    int cnt = 0, cur = 0;
    sort(a + 1, a + n + 1, [&](long long x, long long y) {
      return (x & ((1ll << i) - 1)) > (y & ((1ll << i) - 1));
    });
    for (int j = (1); j <= (n); ++j) cnt += a[j] >> i & 1;
    for (int j = (0); j <= (n); ++j) {
      cur += a[j] >> i & 1;
      int A = cur, B = j - A, C = cnt - cur, D = n - j - C;
      umn(dp[i + 1][A + B + C], dp[i][j] + A + D);
      umn(dp[i + 1][A], dp[i][j] + B + C);
    }
  }
  printf("%d\n", dp[61][0]);
  return 0;
}

```