#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, dp[63][N], s[N], num;
long long a[N];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline bool cmp(long long x, long long y) {
  return (x & ((1LL << num) - 1)) > (y & ((1ll << num) - 1));
}
int main() {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); ++i) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int i = (int)(1); i <= (int)(n); ++i) a[i] = a[n] - a[i];
  memset(dp, 0x3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = (int)(1); i <= (int)(60); ++i) {
    for (int j = (int)(1); j <= (int)(n); ++j)
      s[j] = s[j - 1] + (a[j] >> (i - 1) & 1);
    for (int j = (int)(0); j <= (int)(n); ++j) {
      if (dp[i - 1][j] >= 0x3f3f3f3f) continue;
      int cnt = j - s[j] + s[n] - s[j], k = s[j];
      dp[i][k] = min(dp[i][k], dp[i - 1][j] + cnt);
      cnt = n - cnt;
      k = j + s[n] - s[j];
      dp[i][k] = min(dp[i][k], dp[i - 1][j] + cnt);
    }
    num = i;
    sort(a + 1, a + n + 1, cmp);
  }
  printf("%\d\n", dp[60][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, dp[63][N], s[N], num;
long long a[N];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline bool cmp(long long x, long long y) {
  return (x & ((1LL << num) - 1)) > (y & ((1ll << num) - 1));
}
int main() {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); ++i) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int i = (int)(1); i <= (int)(n); ++i) a[i] = a[n] - a[i];
  memset(dp, 0x3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = (int)(1); i <= (int)(60); ++i) {
    for (int j = (int)(1); j <= (int)(n); ++j)
      s[j] = s[j - 1] + (a[j] >> (i - 1) & 1);
    for (int j = (int)(0); j <= (int)(n); ++j) {
      if (dp[i - 1][j] >= 0x3f3f3f3f) continue;
      int cnt = j - s[j] + s[n] - s[j], k = s[j];
      dp[i][k] = min(dp[i][k], dp[i - 1][j] + cnt);
      cnt = n - cnt;
      k = j + s[n] - s[j];
      dp[i][k] = min(dp[i][k], dp[i - 1][j] + cnt);
    }
    num = i;
    sort(a + 1, a + n + 1, cmp);
  }
  printf("%\d\n", dp[60][0]);
  return 0;
}

```