#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const int N = 100010;
int n, dp[61][N];
struct dat {
  long long a, b;
} a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  scanf("%d", &n);
  long long ma = 0;
  for (int i = (1); i <= (n); ++i) scanf("%lld", &a[i].a), ma = max(ma, a[i].a);
  for (int i = (1); i <= (n); ++i) a[i].a = ma - a[i].a;
  for (int i = (0); i <= (60); ++i)
    for (int j = (0); j <= (n); ++j) dp[i][j] = 1e8;
  dp[0][0] = 0;
  for (int i = (1); i <= (60); ++i) {
    long long tmp = (1ll << i - 1) - 1;
    for (int j = (1); j <= (n); ++j) a[j].b = a[j].a & tmp;
    sort(a + 1, a + n + 1, [](dat x, dat y) { return x.b > y.b; });
    int num1 = 0, cnt = 0;
    for (int j = (1); j <= (n); ++j)
      if (a[j].a & (1ll << i - 1)) ++num1;
    for (int j = (0); j <= (n); ++j) {
      if (a[j].a & (1ll << i - 1)) ++cnt;
      dp[i][cnt] = min(dp[i][cnt], num1 + j - 2 * cnt + dp[i - 1][j]);
      dp[i][num1 + j - cnt] =
          min(dp[i][num1 + j - cnt], n - num1 - j + 2 * cnt + dp[i - 1][j]);
    }
  }
  printf("%d\n", dp[60][0]);
  return 0;
}

### Prompt

Please formulate 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 long long MOD = 1e9 + 7;
const int N = 100010;
int n, dp[61][N];
struct dat {
  long long a, b;
} a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  scanf("%d", &n);
  long long ma = 0;
  for (int i = (1); i <= (n); ++i) scanf("%lld", &a[i].a), ma = max(ma, a[i].a);
  for (int i = (1); i <= (n); ++i) a[i].a = ma - a[i].a;
  for (int i = (0); i <= (60); ++i)
    for (int j = (0); j <= (n); ++j) dp[i][j] = 1e8;
  dp[0][0] = 0;
  for (int i = (1); i <= (60); ++i) {
    long long tmp = (1ll << i - 1) - 1;
    for (int j = (1); j <= (n); ++j) a[j].b = a[j].a & tmp;
    sort(a + 1, a + n + 1, [](dat x, dat y) { return x.b > y.b; });
    int num1 = 0, cnt = 0;
    for (int j = (1); j <= (n); ++j)
      if (a[j].a & (1ll << i - 1)) ++num1;
    for (int j = (0); j <= (n); ++j) {
      if (a[j].a & (1ll << i - 1)) ++cnt;
      dp[i][cnt] = min(dp[i][cnt], num1 + j - 2 * cnt + dp[i - 1][j]);
      dp[i][num1 + j - cnt] =
          min(dp[i][num1 + j - cnt], n - num1 - j + 2 * cnt + dp[i - 1][j]);
    }
  }
  printf("%d\n", dp[60][0]);
  return 0;
}

```