#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int MAX_N = 1e5;
const int INF = 1e9;
int n;
LL a[MAX_N];
int dp[2][MAX_N + 1];
int main() {
  scanf("%d", &n);
  LL mx = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 0; i < n; ++i) {
    a[i] = mx - a[i];
  }
  for (int i = 1; i <= n; ++i) {
    dp[0][i] = INF;
  }
  int cur = 0;
  for (int k = 1; k <= 62; ++k) {
    cur ^= 1;
    for (int i = 0; i <= n; ++i) {
      dp[cur][i] = INF;
    }
    int tot = 0;
    for (int i = 0; i < n; ++i) {
      if (a[i] >> (k - 1) & 1) {
        ++tot;
      }
    }
    for (int i = 0, j = 0; i <= n; ++i) {
      dp[cur][j] = min(dp[cur][j], dp[cur ^ 1][i] + i - j + tot - j);
      dp[cur][i - j + tot] =
          min(dp[cur][i - j + tot], dp[cur ^ 1][i] + n - tot - i + j + j);
      if (i < n && (a[i] >> (k - 1) & 1)) {
        ++j;
      }
    }
    sort(a, a + n, [&](LL x, LL y) {
      return (x & ((1ll << k) - 1)) > (y & ((1ll << k) - 1));
    });
  }
  int ans = INF;
  for (int i = 0; i <= n; ++i) {
    ans = min(ans, dp[cur][i] + i);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
using LL = long long;
const int MAX_N = 1e5;
const int INF = 1e9;
int n;
LL a[MAX_N];
int dp[2][MAX_N + 1];
int main() {
  scanf("%d", &n);
  LL mx = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = 0; i < n; ++i) {
    a[i] = mx - a[i];
  }
  for (int i = 1; i <= n; ++i) {
    dp[0][i] = INF;
  }
  int cur = 0;
  for (int k = 1; k <= 62; ++k) {
    cur ^= 1;
    for (int i = 0; i <= n; ++i) {
      dp[cur][i] = INF;
    }
    int tot = 0;
    for (int i = 0; i < n; ++i) {
      if (a[i] >> (k - 1) & 1) {
        ++tot;
      }
    }
    for (int i = 0, j = 0; i <= n; ++i) {
      dp[cur][j] = min(dp[cur][j], dp[cur ^ 1][i] + i - j + tot - j);
      dp[cur][i - j + tot] =
          min(dp[cur][i - j + tot], dp[cur ^ 1][i] + n - tot - i + j + j);
      if (i < n && (a[i] >> (k - 1) & 1)) {
        ++j;
      }
    }
    sort(a, a + n, [&](LL x, LL y) {
      return (x & ((1ll << k) - 1)) > (y & ((1ll << k) - 1));
    });
  }
  int ans = INF;
  for (int i = 0; i <= n; ++i) {
    ans = min(ans, dp[cur][i] + i);
  }
  printf("%d\n", ans);
  return 0;
}

```