#include <bits/stdc++.h>
using namespace std;
const int n0 = 1e5 + 123, k0 = 60, inf = 1e9;
int n, ans = inf;
long long a[n0], b[2][n0], mx;
int popc(long long x) { return __builtin_popcountll(x); }
int dp[2][n0];
void solve(int k) {
  for (int i = 0; i < n; i++) b[k & 1][i] = a[i] % (1ll << k);
  b[k & 1][n] = 0;
  b[k & 1][n + 1] = (1ll << (k - 1));
  sort(b[k & 1], b[k & 1] + n + 2);
  int p = (k - 1) & 1, j = 0, one = 0;
  for (int i = 0; i < n + 2; i++) {
    dp[k & 1][i] = inf;
    if (!one && b[k & 1][i] >> (k - 1) & 1) {
      one = 1;
      j = 0;
    }
    long long r = 1ll << k;
    if (i < n + 1) {
      r = b[k & 1][i + 1];
      if (one) r ^= (1ll << (k - 1));
    }
    while (j < n + 2 && b[p][j] < r) {
      dp[k & 1][i] =
          min(dp[k & 1][i], dp[p][j] + one * n + (n + 2 - i - 1 - !one));
      j++;
    }
    if (b[k & 1][i] >= mx) ans = min(ans, dp[k & 1][i]);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  int C = 0;
  for (int i = 0; i < n; i++) cin >> a[i], C -= popc(a[i]), mx = max(mx, a[i]);
  for (int i = 1; i < k0; i++) solve(i);
  cout << ans + C;
}

### Prompt

Your task is to create 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;
const int n0 = 1e5 + 123, k0 = 60, inf = 1e9;
int n, ans = inf;
long long a[n0], b[2][n0], mx;
int popc(long long x) { return __builtin_popcountll(x); }
int dp[2][n0];
void solve(int k) {
  for (int i = 0; i < n; i++) b[k & 1][i] = a[i] % (1ll << k);
  b[k & 1][n] = 0;
  b[k & 1][n + 1] = (1ll << (k - 1));
  sort(b[k & 1], b[k & 1] + n + 2);
  int p = (k - 1) & 1, j = 0, one = 0;
  for (int i = 0; i < n + 2; i++) {
    dp[k & 1][i] = inf;
    if (!one && b[k & 1][i] >> (k - 1) & 1) {
      one = 1;
      j = 0;
    }
    long long r = 1ll << k;
    if (i < n + 1) {
      r = b[k & 1][i + 1];
      if (one) r ^= (1ll << (k - 1));
    }
    while (j < n + 2 && b[p][j] < r) {
      dp[k & 1][i] =
          min(dp[k & 1][i], dp[p][j] + one * n + (n + 2 - i - 1 - !one));
      j++;
    }
    if (b[k & 1][i] >= mx) ans = min(ans, dp[k & 1][i]);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  int C = 0;
  for (int i = 0; i < n; i++) cin >> a[i], C -= popc(a[i]), mx = max(mx, a[i]);
  for (int i = 1; i < k0; i++) solve(i);
  cout << ans + C;
}

```