#include <bits/stdc++.h>
const int MAX_N = int(1e5), LOGN = 60;
const long long MAX_LONG = (1LL << 60);
int n, sum[MAX_N + 5];
long long a[MAX_N + 5], dp[LOGN + 5][MAX_N + 5];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", a + i);
  }
  std::sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    a[i] = a[n] - a[i];
  }
  for (int i = 0; i <= LOGN; ++i) {
    for (int j = 0; j <= n; ++j) {
      dp[i][j] = MAX_LONG;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= LOGN; ++i) {
    for (int j = 1; j <= n; ++j) {
      sum[j] = sum[j - 1] + (a[j] >> (i - 1) & 1);
    }
    for (int j = 0; j <= n; ++j) {
      if (dp[i - 1][j] != MAX_LONG) {
        int cnt = j - sum[j] + sum[n] - sum[j];
        dp[i][sum[j]] = std::min(dp[i][sum[j]], dp[i - 1][j] + cnt);
        cnt = sum[j] + (n - j) - (sum[n] - sum[j]);
        dp[i][j + sum[n] - sum[j]] =
            std::min(dp[i][j + sum[n] - sum[j]], dp[i - 1][j] + cnt);
      }
    }
    std::sort(a + 1, a + n + 1, [&](const long long &x, const long long &y) {
      return x % (1LL << (i)) > y % (1LL << (i));
    });
  }
  printf("%I64d\n", dp[LOGN][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>
const int MAX_N = int(1e5), LOGN = 60;
const long long MAX_LONG = (1LL << 60);
int n, sum[MAX_N + 5];
long long a[MAX_N + 5], dp[LOGN + 5][MAX_N + 5];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", a + i);
  }
  std::sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    a[i] = a[n] - a[i];
  }
  for (int i = 0; i <= LOGN; ++i) {
    for (int j = 0; j <= n; ++j) {
      dp[i][j] = MAX_LONG;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= LOGN; ++i) {
    for (int j = 1; j <= n; ++j) {
      sum[j] = sum[j - 1] + (a[j] >> (i - 1) & 1);
    }
    for (int j = 0; j <= n; ++j) {
      if (dp[i - 1][j] != MAX_LONG) {
        int cnt = j - sum[j] + sum[n] - sum[j];
        dp[i][sum[j]] = std::min(dp[i][sum[j]], dp[i - 1][j] + cnt);
        cnt = sum[j] + (n - j) - (sum[n] - sum[j]);
        dp[i][j + sum[n] - sum[j]] =
            std::min(dp[i][j + sum[n] - sum[j]], dp[i - 1][j] + cnt);
      }
    }
    std::sort(a + 1, a + n + 1, [&](const long long &x, const long long &y) {
      return x % (1LL << (i)) > y % (1LL << (i));
    });
  }
  printf("%I64d\n", dp[LOGN][0]);
}

```