#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, M = 62, INF = 1e9 + 7;
int n, dp[M][N], p[N], t[N];
long long a[N];
int c[2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++)
    if (!(a[i] & 1ll)) ++c[0];
  c[1] = n;
  for (int k = 0; k < M; k++)
    for (int i = 1; i <= n + 1; i++) dp[k][i] = INF;
  dp[0][c[0] + 1] = n - c[0], dp[0][n + 1] = min(dp[0][n + 1], c[0]);
  for (int i = 1; i <= n; i++) p[i] = i;
  for (int k = 0; k < M - 1; k++) {
    for (int i = 1; i <= n; i++) t[i] = p[i];
    c[0] = 0, c[1] = n;
    for (int i = 1; i <= n; i++)
      if (!((a[i] >> k) & 1ll)) ++c[0];
    for (int i = n; i; i--)
      if ((a[t[i]] >> k) & 1ll)
        p[c[1]--] = t[i];
      else
        p[c[0]--] = t[i];
    c[1] = 0;
    for (int i = 1; i <= n; i++)
      if ((a[i] >> (k + 1)) & 1ll) ++c[1];
    int ji = 0;
    for (int i = n + 1; i; i--) {
      dp[k + 1][n - ji + 1] = min(dp[k + 1][n - ji + 1], dp[k][i] + n - c[1]);
      dp[k + 1][n - (ji + c[1]) + 1] =
          min(dp[k + 1][n - (ji + c[1]) + 1], dp[k][i] + c[1]);
      if ((a[p[i - 1]] >> (k + 1)) & 1ll)
        ++ji, --c[1];
      else
        ++c[1];
    }
  }
  printf("%d", dp[M - 1][n + 1]);
  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 = 1e5 + 5, M = 62, INF = 1e9 + 7;
int n, dp[M][N], p[N], t[N];
long long a[N];
int c[2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++)
    if (!(a[i] & 1ll)) ++c[0];
  c[1] = n;
  for (int k = 0; k < M; k++)
    for (int i = 1; i <= n + 1; i++) dp[k][i] = INF;
  dp[0][c[0] + 1] = n - c[0], dp[0][n + 1] = min(dp[0][n + 1], c[0]);
  for (int i = 1; i <= n; i++) p[i] = i;
  for (int k = 0; k < M - 1; k++) {
    for (int i = 1; i <= n; i++) t[i] = p[i];
    c[0] = 0, c[1] = n;
    for (int i = 1; i <= n; i++)
      if (!((a[i] >> k) & 1ll)) ++c[0];
    for (int i = n; i; i--)
      if ((a[t[i]] >> k) & 1ll)
        p[c[1]--] = t[i];
      else
        p[c[0]--] = t[i];
    c[1] = 0;
    for (int i = 1; i <= n; i++)
      if ((a[i] >> (k + 1)) & 1ll) ++c[1];
    int ji = 0;
    for (int i = n + 1; i; i--) {
      dp[k + 1][n - ji + 1] = min(dp[k + 1][n - ji + 1], dp[k][i] + n - c[1]);
      dp[k + 1][n - (ji + c[1]) + 1] =
          min(dp[k + 1][n - (ji + c[1]) + 1], dp[k][i] + c[1]);
      if ((a[p[i - 1]] >> (k + 1)) & 1ll)
        ++ji, --c[1];
      else
        ++c[1];
    }
  }
  printf("%d", dp[M - 1][n + 1]);
  return 0;
}

```