#include <bits/stdc++.h>
using namespace std;
int n, D;
long long a[100012], dp[66][100012];
bool cmp(long long x, long long y) {
  return (x & ((1ll << D) - 1)) > (y & ((1ll << D) - 1));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  int ct = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] & 1) ct++;
  for (int i = 0; i <= 60; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 1e18;
  dp[0][ct] = n - ct;
  dp[0][0] = ct;
  for (int i = 1; i <= 60; i++) {
    D = i;
    sort(a + 1, a + 1 + n, cmp);
    int s0 = 0, s1 = 0, ss0 = 0, ss1 = 0;
    for (int j = 1; j <= n; j++)
      if (a[j] & (1ll << i)) ss1++;
    ss0 = n - ss1;
    for (int j = 0; j <= n; j++) {
      dp[i][s1] = min(dp[i][s1], dp[i - 1][j] + ss1 + s0);
      dp[i][j + ss1] = min(dp[i][j + ss1], dp[i - 1][j] + ss0 + s1);
      if (a[j + 1] & (1ll << i))
        ss1--, s1++;
      else
        ss0--, s0++;
    }
  }
  printf("%lld\n", dp[60][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;
int n, D;
long long a[100012], dp[66][100012];
bool cmp(long long x, long long y) {
  return (x & ((1ll << D) - 1)) > (y & ((1ll << D) - 1));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i];
  int ct = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] & 1) ct++;
  for (int i = 0; i <= 60; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 1e18;
  dp[0][ct] = n - ct;
  dp[0][0] = ct;
  for (int i = 1; i <= 60; i++) {
    D = i;
    sort(a + 1, a + 1 + n, cmp);
    int s0 = 0, s1 = 0, ss0 = 0, ss1 = 0;
    for (int j = 1; j <= n; j++)
      if (a[j] & (1ll << i)) ss1++;
    ss0 = n - ss1;
    for (int j = 0; j <= n; j++) {
      dp[i][s1] = min(dp[i][s1], dp[i - 1][j] + ss1 + s0);
      dp[i][j + ss1] = min(dp[i][j + ss1], dp[i - 1][j] + ss0 + s1);
      if (a[j + 1] & (1ll << i))
        ss1--, s1++;
      else
        ss0--, s0++;
    }
  }
  printf("%lld\n", dp[60][0]);
}

```