#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
const int M = 63;
long long a[N];
int b[N], f[M][N], g[2][N], now;
bool cmp(int x, int y) {
  return (a[x] & ((1LL << now) - 1)) < (a[y] & ((1LL << now) - 1));
}
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - 48;
    c = getchar();
  }
  return x * f;
}
signed main() {
  int n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; ++i) a[i] = a[n] - a[i], b[i] = i;
  memset(f, 0x3f, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k <= M - 2; ++k) {
    now = k;
    sort(b + 1, b + 1 + n, cmp);
    for (int i = 1; i <= n; ++i) {
      g[0][i] = g[0][i - 1];
      g[1][i] = g[1][i - 1];
      g[(a[b[i]] >> k) & 1][i]++;
    }
    for (int i = 0; i <= n; ++i) {
      f[k + 1][g[1][n] - g[1][n - i]] =
          min(f[k + 1][g[1][n] - g[1][n - i]],
              f[k][i] + g[1][n - i] + g[0][n] - g[0][n - i]);
      f[k + 1][n - g[0][n - i]] =
          min(f[k + 1][n - g[0][n - i]],
              f[k][i] + g[0][n - i] + g[1][n] - g[1][n - i]);
    }
  }
  cout << f[M - 1][0] << endl;
  fclose(stdin);
  fclose(stdout);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int M = 63;
long long a[N];
int b[N], f[M][N], g[2][N], now;
bool cmp(int x, int y) {
  return (a[x] & ((1LL << now) - 1)) < (a[y] & ((1LL << now) - 1));
}
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - 48;
    c = getchar();
  }
  return x * f;
}
signed main() {
  int n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; ++i) a[i] = a[n] - a[i], b[i] = i;
  memset(f, 0x3f, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k <= M - 2; ++k) {
    now = k;
    sort(b + 1, b + 1 + n, cmp);
    for (int i = 1; i <= n; ++i) {
      g[0][i] = g[0][i - 1];
      g[1][i] = g[1][i - 1];
      g[(a[b[i]] >> k) & 1][i]++;
    }
    for (int i = 0; i <= n; ++i) {
      f[k + 1][g[1][n] - g[1][n - i]] =
          min(f[k + 1][g[1][n] - g[1][n - i]],
              f[k][i] + g[1][n - i] + g[0][n] - g[0][n - i]);
      f[k + 1][n - g[0][n - i]] =
          min(f[k + 1][n - g[0][n - i]],
              f[k][i] + g[0][n - i] + g[1][n] - g[1][n - i]);
    }
  }
  cout << f[M - 1][0] << endl;
  fclose(stdin);
  fclose(stdout);
  return 0;
}

```