#include <bits/stdc++.h>
using namespace std;
long long dp[65][100005], a[100005], mx, ans;
int n, sum0[100005], sum1[100005], k[65][100005];
void read(long long &x) {
  char c = getchar();
  x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
}
int main() {
  scanf("%d", &n);
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 1; i <= n; i++) read(a[i]), mx = max(mx, a[i]);
  for (int i = 1; i <= n; i++) a[i] = mx - a[i];
  dp[0][0] = 0;
  for (int i = 0; i <= 59; i++) {
    if (i == 0)
      for (int j = 1; j <= n; j++) k[i][j] = j;
    else {
      int cnt = 0;
      for (int j = 1; j <= n; j++)
        if (!((a[k[i - 1][j]] >> (i - 1)) & 1)) k[i][++cnt] = k[i - 1][j];
      for (int j = 1; j <= n; j++)
        if ((a[k[i - 1][j]] >> (i - 1)) & 1) k[i][++cnt] = k[i - 1][j];
    }
    for (int j = 1; j <= n; j++) {
      sum0[j] = sum0[j - 1], sum1[j] = sum1[j - 1];
      if ((a[k[i][j]] >> i) & 1)
        sum1[j]++;
      else
        sum0[j]++;
    }
    int val, sta;
    for (int j = 0; j <= n; j++) {
      val = sum1[n - j] + sum0[n] - sum0[n - j], sta = sum1[n] - sum1[n - j];
      dp[i + 1][sta] = min(dp[i + 1][sta], dp[i][j] + val);
      val = sum0[n - j] + sum1[n] - sum1[n - j], sta = n - sum0[n - j];
      dp[i + 1][sta] = min(dp[i + 1][sta], dp[i][j] + val);
    }
  }
  printf("%lld\n", dp[60][0]);
  return 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;
long long dp[65][100005], a[100005], mx, ans;
int n, sum0[100005], sum1[100005], k[65][100005];
void read(long long &x) {
  char c = getchar();
  x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
}
int main() {
  scanf("%d", &n);
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 1; i <= n; i++) read(a[i]), mx = max(mx, a[i]);
  for (int i = 1; i <= n; i++) a[i] = mx - a[i];
  dp[0][0] = 0;
  for (int i = 0; i <= 59; i++) {
    if (i == 0)
      for (int j = 1; j <= n; j++) k[i][j] = j;
    else {
      int cnt = 0;
      for (int j = 1; j <= n; j++)
        if (!((a[k[i - 1][j]] >> (i - 1)) & 1)) k[i][++cnt] = k[i - 1][j];
      for (int j = 1; j <= n; j++)
        if ((a[k[i - 1][j]] >> (i - 1)) & 1) k[i][++cnt] = k[i - 1][j];
    }
    for (int j = 1; j <= n; j++) {
      sum0[j] = sum0[j - 1], sum1[j] = sum1[j - 1];
      if ((a[k[i][j]] >> i) & 1)
        sum1[j]++;
      else
        sum0[j]++;
    }
    int val, sta;
    for (int j = 0; j <= n; j++) {
      val = sum1[n - j] + sum0[n] - sum0[n - j], sta = sum1[n] - sum1[n - j];
      dp[i + 1][sta] = min(dp[i + 1][sta], dp[i][j] + val);
      val = sum0[n - j] + sum1[n] - sum1[n - j], sta = n - sum0[n - j];
      dp[i + 1][sta] = min(dp[i + 1][sta], dp[i][j] + val);
    }
  }
  printf("%lld\n", dp[60][0]);
  return 0;
}

```