#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int n, id[maxn], f[maxn], g[maxn], s[maxn];
long long a[maxn], b[maxn];
void chk(int &x, int y) {
  if (x > y) x = y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    a[i] = a[n] - a[i];
  }
  memset(f, 0x3f, sizeof(f)), f[0] = 0;
  for (int i = 0; i <= 57; i++, swap(f, g)) {
    memset(g, 0x3f, sizeof(g));
    for (int j = 1; j <= n; j++) {
      b[j] = a[j] & ((1LL << i) - 1), id[j] = j;
    }
    sort(id + 1, id + n + 1, [&](int x, int y) { return b[x] > b[y]; });
    for (int j = 1; j <= n; j++) {
      s[j] = s[j - 1] + (a[id[j]] >> i & 1);
    }
    for (int j = 0; j <= n; j++) {
      chk(g[s[j]], f[j] + j - s[j] + s[n] - s[j]);
      chk(g[j + s[n] - s[j]], f[j] + s[j] + n - j - (s[n] - s[j]));
    }
  }
  printf("%d\n", f[0]);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 maxn = 100010;
int n, id[maxn], f[maxn], g[maxn], s[maxn];
long long a[maxn], b[maxn];
void chk(int &x, int y) {
  if (x > y) x = y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    a[i] = a[n] - a[i];
  }
  memset(f, 0x3f, sizeof(f)), f[0] = 0;
  for (int i = 0; i <= 57; i++, swap(f, g)) {
    memset(g, 0x3f, sizeof(g));
    for (int j = 1; j <= n; j++) {
      b[j] = a[j] & ((1LL << i) - 1), id[j] = j;
    }
    sort(id + 1, id + n + 1, [&](int x, int y) { return b[x] > b[y]; });
    for (int j = 1; j <= n; j++) {
      s[j] = s[j - 1] + (a[id[j]] >> i & 1);
    }
    for (int j = 0; j <= n; j++) {
      chk(g[s[j]], f[j] + j - s[j] + s[n] - s[j]);
      chk(g[j + s[n] - s[j]], f[j] + s[j] + n - j - (s[n] - s[j]));
    }
  }
  printf("%d\n", f[0]);
  return 0;
}

```