#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, t0, t1, s0[N], s1[N];
long long a[N], f[N], g[N], dp[70][N];
inline void srt(long long x) {
  int cc = 0;
  for (int i = 1; i <= n; ++i)
    if (!(a[f[i]] >> x & 1)) g[++cc] = f[i];
  for (int i = 1; i <= n; ++i)
    if (a[f[i]] >> x & 1) g[++cc] = f[i];
  for (int i = 1; i <= n; ++i) f[i] = g[i];
}
inline void init(long long x) {
  for (int i = 1; i <= n; ++i) {
    s0[i] = s0[i - 1];
    s1[i] = s1[i - 1];
    if (a[f[i]] >> x & 1)
      s1[i]++;
    else
      s0[i]++;
  }
  t1 = s1[n];
  t0 = s0[n];
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]), mx = max(mx, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = mx - a[i], f[i] = i;
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (long long i = 0; i < 62; ++i) {
    if (i) srt(i - 1);
    init(i);
    for (int j = 0; j <= n; ++j) {
      int w, stt;
      w = s1[n - j] + t0 - s0[n - j];
      stt = t1 - s1[n - j];
      dp[i + 1][stt] = min(dp[i + 1][stt], dp[i][j] + w);
      w = s0[n - j] + t1 - s1[n - j];
      stt = n - s0[n - j];
      dp[i + 1][stt] = min(dp[i + 1][stt], dp[i][j] + w);
    }
  }
  printf("%lld\n", dp[62][0]);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 100010;
int n, t0, t1, s0[N], s1[N];
long long a[N], f[N], g[N], dp[70][N];
inline void srt(long long x) {
  int cc = 0;
  for (int i = 1; i <= n; ++i)
    if (!(a[f[i]] >> x & 1)) g[++cc] = f[i];
  for (int i = 1; i <= n; ++i)
    if (a[f[i]] >> x & 1) g[++cc] = f[i];
  for (int i = 1; i <= n; ++i) f[i] = g[i];
}
inline void init(long long x) {
  for (int i = 1; i <= n; ++i) {
    s0[i] = s0[i - 1];
    s1[i] = s1[i - 1];
    if (a[f[i]] >> x & 1)
      s1[i]++;
    else
      s0[i]++;
  }
  t1 = s1[n];
  t0 = s0[n];
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]), mx = max(mx, a[i]);
  for (int i = 1; i <= n; ++i) a[i] = mx - a[i], f[i] = i;
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (long long i = 0; i < 62; ++i) {
    if (i) srt(i - 1);
    init(i);
    for (int j = 0; j <= n; ++j) {
      int w, stt;
      w = s1[n - j] + t0 - s0[n - j];
      stt = t1 - s1[n - j];
      dp[i + 1][stt] = min(dp[i + 1][stt], dp[i][j] + w);
      w = s0[n - j] + t1 - s1[n - j];
      stt = n - s0[n - j];
      dp[i + 1][stt] = min(dp[i + 1][stt], dp[i][j] + w);
    }
  }
  printf("%lld\n", dp[62][0]);
  return 0;
}

```