#include <bits/stdc++.h>
using namespace std;
long long int a[100000];
int dp[60][100001];
int r;
bool comp(long long int a, long long int b) {
  return (a & ((1LL << r) - 1)) > (b & ((1LL << r) - 1));
}
int main() {
  int i;
  int n;
  long long int m = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%I64d", &a[i]), m = max(m, a[i]);
  for (i = 0; i < n; i++) a[i] = m - a[i];
  int j;
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) dp[0][i] = 1e9;
  for (i = 0; i < 59; i++) {
    r = i;
    sort(a, a + n, comp);
    for (j = 0; j <= n; j++) dp[i + 1][j] = 1e9;
    int c = 0, b = 0;
    for (j = 0; j < n; j++) {
      if (a[j] & (1LL << i)) b++;
    }
    for (j = 0; j <= n; j++) {
      dp[i + 1][c] = min(dp[i + 1][c], dp[i][j] + b + j - 2 * c);
      dp[i + 1][j + b - c] =
          min(dp[i + 1][j + b - c], dp[i][j] + n - (b + j - 2 * c));
      if ((j < n) && (a[j] & (1LL << i))) c++;
    }
  }
  printf("%d\n", dp[59][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;
long long int a[100000];
int dp[60][100001];
int r;
bool comp(long long int a, long long int b) {
  return (a & ((1LL << r) - 1)) > (b & ((1LL << r) - 1));
}
int main() {
  int i;
  int n;
  long long int m = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%I64d", &a[i]), m = max(m, a[i]);
  for (i = 0; i < n; i++) a[i] = m - a[i];
  int j;
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) dp[0][i] = 1e9;
  for (i = 0; i < 59; i++) {
    r = i;
    sort(a, a + n, comp);
    for (j = 0; j <= n; j++) dp[i + 1][j] = 1e9;
    int c = 0, b = 0;
    for (j = 0; j < n; j++) {
      if (a[j] & (1LL << i)) b++;
    }
    for (j = 0; j <= n; j++) {
      dp[i + 1][c] = min(dp[i + 1][c], dp[i][j] + b + j - 2 * c);
      dp[i + 1][j + b - c] =
          min(dp[i + 1][j + b - c], dp[i][j] + n - (b + j - 2 * c));
      if ((j < n) && (a[j] & (1LL << i))) c++;
    }
  }
  printf("%d\n", dp[59][0]);
  return 0;
}

```