#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
const int INF = 0x3f3f3f3f;
int n, dp[61][maxn];
long long a1[maxn], a2[maxn], a[maxn];
int s0[maxn], s1[maxn];
inline void init(int k) {
  if (k >= 0) {
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] & (1LL << k))
        a2[++cnt2] = a[i];
      else
        a1[++cnt1] = a[i];
    }
    for (int i = 1; i <= cnt1; i++) a[i] = a1[i];
    for (int i = 1; i <= cnt2; i++) a[i + cnt1] = a2[i];
  }
  s0[0] = s1[0] = 0;
  for (int i = 1; i <= n; i++) {
    s0[i] = s0[i - 1];
    s1[i] = s1[i - 1];
    if (a[i] & (1LL << (k + 1)))
      s1[i]++;
    else
      s0[i]++;
  }
}
inline void upmin(int &x, int y) { x = min(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(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i <= 58; i++) {
    init(i - 1);
    for (int j = 0; j <= n; j++) {
      int w = s1[n - j] + s0[n] - s0[n - j];
      int nxt = s1[n] - s1[n - j];
      upmin(dp[i + 1][nxt], dp[i][j] + w);
      w = s0[n - j] + s1[n] - s1[n - j];
      nxt = s1[n - j] + j;
      upmin(dp[i + 1][nxt], dp[i][j] + w);
    }
  }
  int Ans = INF;
  for (int i = 0; i <= n; i++) {
    upmin(Ans, dp[59][i] + i);
  }
  printf("%d\n", Ans);
  return 0;
}

### Prompt

Generate 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 maxn = 100000 + 10;
const int INF = 0x3f3f3f3f;
int n, dp[61][maxn];
long long a1[maxn], a2[maxn], a[maxn];
int s0[maxn], s1[maxn];
inline void init(int k) {
  if (k >= 0) {
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] & (1LL << k))
        a2[++cnt2] = a[i];
      else
        a1[++cnt1] = a[i];
    }
    for (int i = 1; i <= cnt1; i++) a[i] = a1[i];
    for (int i = 1; i <= cnt2; i++) a[i + cnt1] = a2[i];
  }
  s0[0] = s1[0] = 0;
  for (int i = 1; i <= n; i++) {
    s0[i] = s0[i - 1];
    s1[i] = s1[i - 1];
    if (a[i] & (1LL << (k + 1)))
      s1[i]++;
    else
      s0[i]++;
  }
}
inline void upmin(int &x, int y) { x = min(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(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i <= 58; i++) {
    init(i - 1);
    for (int j = 0; j <= n; j++) {
      int w = s1[n - j] + s0[n] - s0[n - j];
      int nxt = s1[n] - s1[n - j];
      upmin(dp[i + 1][nxt], dp[i][j] + w);
      w = s0[n - j] + s1[n] - s1[n - j];
      nxt = s1[n - j] + j;
      upmin(dp[i + 1][nxt], dp[i][j] + w);
    }
  }
  int Ans = INF;
  for (int i = 0; i <= n; i++) {
    upmin(Ans, dp[59][i] + i);
  }
  printf("%d\n", Ans);
  return 0;
}

```