#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
long long dp[70][maxn];
long long a[maxn], f[maxn], f1[maxn];
int cnt;
int sum0[maxn], sum1[maxn], tot0, tot1, n;
void Sort(long long x) {
  cnt = 0;
  for (int i = 1; i <= n; i++)
    if (((a[f[i]] >> x) & 1) == 0) f1[++cnt] = f[i];
  for (int i = 1; i <= n; i++)
    if ((a[f[i]] >> x) & 1) f1[++cnt] = f[i];
  for (int i = 1; i <= n; i++) f[i] = f1[i];
}
void init(long long x) {
  for (int i = 1; i <= n; i++) {
    sum0[i] = sum0[i - 1];
    sum1[i] = sum1[i - 1];
    if ((a[f[i]] >> x) & 1)
      sum1[i]++;
    else
      sum0[i]++;
  }
  tot1 = sum1[n], tot0 = sum0[n];
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = mx - a[i];
  }
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 1; i <= n; i++) f[i] = i;
  dp[0][0] = 0;
  for (long long i = 0; i < 62; i++) {
    if (i != 0) {
      Sort(i - 1);
    }
    init(i);
    for (int j = 0; j <= n; j++) {
      for (int bit = 0; bit < 2; bit++) {
        if (bit == 0) {
          int val = sum1[n - j] + tot0 - sum0[n - j],
              Next_state = tot1 - sum1[n - j];
          dp[i + 1][Next_state] = min(dp[i + 1][Next_state], dp[i][j] + val);
        } else if (bit == 1) {
          int val = sum0[n - j] + tot1 - sum1[n - j],
              Next_state = n - sum0[n - j];
          dp[i + 1][Next_state] = min(dp[i + 1][Next_state], dp[i][j] + val);
        }
      }
    }
  }
  printf("%lld\n", dp[62][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;
const int maxn = 100010;
long long dp[70][maxn];
long long a[maxn], f[maxn], f1[maxn];
int cnt;
int sum0[maxn], sum1[maxn], tot0, tot1, n;
void Sort(long long x) {
  cnt = 0;
  for (int i = 1; i <= n; i++)
    if (((a[f[i]] >> x) & 1) == 0) f1[++cnt] = f[i];
  for (int i = 1; i <= n; i++)
    if ((a[f[i]] >> x) & 1) f1[++cnt] = f[i];
  for (int i = 1; i <= n; i++) f[i] = f1[i];
}
void init(long long x) {
  for (int i = 1; i <= n; i++) {
    sum0[i] = sum0[i - 1];
    sum1[i] = sum1[i - 1];
    if ((a[f[i]] >> x) & 1)
      sum1[i]++;
    else
      sum0[i]++;
  }
  tot1 = sum1[n], tot0 = sum0[n];
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = mx - a[i];
  }
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 1; i <= n; i++) f[i] = i;
  dp[0][0] = 0;
  for (long long i = 0; i < 62; i++) {
    if (i != 0) {
      Sort(i - 1);
    }
    init(i);
    for (int j = 0; j <= n; j++) {
      for (int bit = 0; bit < 2; bit++) {
        if (bit == 0) {
          int val = sum1[n - j] + tot0 - sum0[n - j],
              Next_state = tot1 - sum1[n - j];
          dp[i + 1][Next_state] = min(dp[i + 1][Next_state], dp[i][j] + val);
        } else if (bit == 1) {
          int val = sum0[n - j] + tot1 - sum1[n - j],
              Next_state = n - sum0[n - j];
          dp[i + 1][Next_state] = min(dp[i + 1][Next_state], dp[i][j] + val);
        }
      }
    }
  }
  printf("%lld\n", dp[62][0]);
}

```