#include <bits/stdc++.h>
using namespace std;
const int N = 100009;
int n, pos[N], sum[N][2], f[63][N], now;
long long m, a[N];
inline bool cmp(int x, int y) {
  return (a[x] & ((1LL << now) - 1)) < (a[y] & ((1LL << now) - 1));
}
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], pos[i] = i;
  memset(f, 0x3f, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k <= 60; ++k) {
    now = k, sort(pos + 1, pos + n + 1, cmp);
    for (int i = 1; i <= n; ++i) {
      sum[i][0] = sum[i - 1][0], sum[i][1] = sum[i - 1][1];
      ++sum[i][(a[pos[i]] >> k) & 1];
    }
    for (int i = 0; i <= n; ++i) {
      int unup0 = sum[n - i][0], unup1 = sum[n - i][1],
          up0 = sum[n][0] - sum[n - i][0], up1 = sum[n][1] - sum[n - i][1];
      f[k + 1][up1] = min(f[k + 1][up1], f[k][i] + up0 + unup1);
      f[k + 1][n - unup0] = min(f[k + 1][n - unup0], f[k][i] + unup0 + up1);
    }
  }
  printf("%d\n", f[61][0]);
  return 0;
}

### Prompt

Your task is to create a CPP solution 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>
using namespace std;
const int N = 100009;
int n, pos[N], sum[N][2], f[63][N], now;
long long m, a[N];
inline bool cmp(int x, int y) {
  return (a[x] & ((1LL << now) - 1)) < (a[y] & ((1LL << now) - 1));
}
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], pos[i] = i;
  memset(f, 0x3f, sizeof(f));
  f[0][0] = 0;
  for (int k = 0; k <= 60; ++k) {
    now = k, sort(pos + 1, pos + n + 1, cmp);
    for (int i = 1; i <= n; ++i) {
      sum[i][0] = sum[i - 1][0], sum[i][1] = sum[i - 1][1];
      ++sum[i][(a[pos[i]] >> k) & 1];
    }
    for (int i = 0; i <= n; ++i) {
      int unup0 = sum[n - i][0], unup1 = sum[n - i][1],
          up0 = sum[n][0] - sum[n - i][0], up1 = sum[n][1] - sum[n - i][1];
      f[k + 1][up1] = min(f[k + 1][up1], f[k][i] + up0 + unup1);
      f[k + 1][n - unup0] = min(f[k + 1][n - unup0], f[k][i] + unup0 + up1);
    }
  }
  printf("%d\n", f[61][0]);
  return 0;
}

```