#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7, K = 135, INF = 1e9;
int dp[K][N];
long long a[N];
long long get(long long x, int i) {
  if (i >= 60) return x;
  return x & ((1LL << (i - 1)) - 1);
}
long long geti(long long x, int i) {
  if (i >= 60) return 0;
  return x & (1LL << (i - 1));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int k = 0; k < K; k++)
    for (int i = 0; i < N; i++) dp[k][i] = INF;
  dp[0][0] = 0;
  int n;
  cin >> n;
  long long mx = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) a[i] = mx - a[i];
  for (int k = 1; k < K; k++) {
    vector<int> order;
    for (int i = 1; i <= n; i++) order.push_back(i);
    sort(order.begin(), order.end(),
         [&k](int i, int j) { return get(a[i], k) > get(a[j], k); });
    for (int b = 0; b <= 1; b++) {
      int cnt = 0, carry = 0;
      for (int i = 1; i <= n; i++) {
        bool bi = geti(a[i], k);
        if (bi + b >= 2) carry++;
        if (bi ^ b) cnt++;
      }
      dp[k][carry] = min(dp[k][carry], dp[k - 1][0] + cnt);
      for (int pc = 1; pc <= n; pc++) {
        int id = order[pc - 1];
        bool bi = geti(a[id], k);
        if (bi + b >= 2) carry--;
        if (bi ^ b) cnt--;
        if (bi ^ b ^ 1) cnt++;
        if (b + bi + 1 >= 2) carry++;
        dp[k][carry] = min(dp[k][carry], dp[k - 1][pc] + cnt);
      }
    }
  }
  int ans = INF;
  for (int i = 0; i <= n; i++) {
    ans = min(ans, dp[K - 1][i]);
  }
  cout << ans << endl;
}

### Prompt

Generate 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 N = 1e5 + 7, K = 135, INF = 1e9;
int dp[K][N];
long long a[N];
long long get(long long x, int i) {
  if (i >= 60) return x;
  return x & ((1LL << (i - 1)) - 1);
}
long long geti(long long x, int i) {
  if (i >= 60) return 0;
  return x & (1LL << (i - 1));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int k = 0; k < K; k++)
    for (int i = 0; i < N; i++) dp[k][i] = INF;
  dp[0][0] = 0;
  int n;
  cin >> n;
  long long mx = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) a[i] = mx - a[i];
  for (int k = 1; k < K; k++) {
    vector<int> order;
    for (int i = 1; i <= n; i++) order.push_back(i);
    sort(order.begin(), order.end(),
         [&k](int i, int j) { return get(a[i], k) > get(a[j], k); });
    for (int b = 0; b <= 1; b++) {
      int cnt = 0, carry = 0;
      for (int i = 1; i <= n; i++) {
        bool bi = geti(a[i], k);
        if (bi + b >= 2) carry++;
        if (bi ^ b) cnt++;
      }
      dp[k][carry] = min(dp[k][carry], dp[k - 1][0] + cnt);
      for (int pc = 1; pc <= n; pc++) {
        int id = order[pc - 1];
        bool bi = geti(a[id], k);
        if (bi + b >= 2) carry--;
        if (bi ^ b) cnt--;
        if (bi ^ b ^ 1) cnt++;
        if (b + bi + 1 >= 2) carry++;
        dp[k][carry] = min(dp[k][carry], dp[k - 1][pc] + cnt);
      }
    }
  }
  int ans = INF;
  for (int i = 0; i <= n; i++) {
    ans = min(ans, dp[K - 1][i]);
  }
  cout << ans << endl;
}

```