#include <bits/stdc++.h>
const int kMaxn = 1e5 + 10;
typedef int IAr[kMaxn];
int n;
IAr DP[60], sum0, sum1, id;
long long A[kMaxn];
template <typename T>
inline void UMin(T& x, const T& y) {
  if (x > y) x = y;
}
void Sort(int);
int main() {
  scanf("%d", &n);
  long long maxa = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", A + i);
    maxa = std::max(maxa, A[i]);
  }
  for (int i = 1; i <= n; i++) {
    A[i] = maxa - A[i], id[i] = i;
  }
  memset(DP, 0x3f, sizeof DP);
  DP[0][0] = 0;
  for (int i = 0; i < 59; i++) {
    if (i) Sort(i - 1);
    for (int j = 1; j <= n; j++) {
      sum0[j] = sum0[j - 1] + (A[id[j]] >> i & 1 ^ 1);
      sum1[j] = sum1[j - 1] + (A[id[j]] >> i & 1);
    }
    for (int j = 0; j <= n; j++) {
      UMin(DP[i + 1][sum1[n] - sum1[n - j]],
           DP[i][j] + sum1[n - j] + sum0[n] - sum0[n - j]);
      UMin(DP[i + 1][n - sum0[n - j]],
           DP[i][j] + sum0[n - j] + sum1[n] - sum1[n - j]);
    }
  }
  printf("%d\n", DP[59][0]);
  return 0;
}
void Sort(int k) {
  static IAr buf;
  int back = 0;
  for (int i = 1; i <= n; i++) {
    if (A[id[i]] >> k & 1 ^ 1) buf[++back] = id[i];
  }
  for (int i = 1; i <= n; i++) {
    if (A[id[i]] >> k & 1) buf[++back] = id[i];
  }
  for (int i = 1; i <= n; i++) id[i] = buf[i];
}

### Prompt

Please create a solution in Cpp 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>
const int kMaxn = 1e5 + 10;
typedef int IAr[kMaxn];
int n;
IAr DP[60], sum0, sum1, id;
long long A[kMaxn];
template <typename T>
inline void UMin(T& x, const T& y) {
  if (x > y) x = y;
}
void Sort(int);
int main() {
  scanf("%d", &n);
  long long maxa = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", A + i);
    maxa = std::max(maxa, A[i]);
  }
  for (int i = 1; i <= n; i++) {
    A[i] = maxa - A[i], id[i] = i;
  }
  memset(DP, 0x3f, sizeof DP);
  DP[0][0] = 0;
  for (int i = 0; i < 59; i++) {
    if (i) Sort(i - 1);
    for (int j = 1; j <= n; j++) {
      sum0[j] = sum0[j - 1] + (A[id[j]] >> i & 1 ^ 1);
      sum1[j] = sum1[j - 1] + (A[id[j]] >> i & 1);
    }
    for (int j = 0; j <= n; j++) {
      UMin(DP[i + 1][sum1[n] - sum1[n - j]],
           DP[i][j] + sum1[n - j] + sum0[n] - sum0[n - j]);
      UMin(DP[i + 1][n - sum0[n - j]],
           DP[i][j] + sum0[n - j] + sum1[n] - sum1[n - j]);
    }
  }
  printf("%d\n", DP[59][0]);
  return 0;
}
void Sort(int k) {
  static IAr buf;
  int back = 0;
  for (int i = 1; i <= n; i++) {
    if (A[id[i]] >> k & 1 ^ 1) buf[++back] = id[i];
  }
  for (int i = 1; i <= n; i++) {
    if (A[id[i]] >> k & 1) buf[++back] = id[i];
  }
  for (int i = 1; i <= n; i++) id[i] = buf[i];
}

```