#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (!(c ^ 45)) f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
int n, dp[65][100005], id[100005], now, s[2][100005];
long long a[100005];
inline char cmp(int x, int y) {
  return (a[x] & ((1ll << now) - 1)) < (a[y] & ((1ll << now) - 1));
}
signed main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i], id[i] = i;
  memset(dp, 0x3f, sizeof(dp)), dp[0][0] = 0;
  for (now = 0; now <= 60; now++) {
    sort(id + 1, id + n + 1, cmp);
    for (int i = 1; i <= n; i++)
      s[0][i] = s[0][i - 1], s[1][i] = s[1][i - 1],
      s[(a[id[i]] >> now) & 1][i]++;
    for (int i = 0; i <= n; i++) {
      int qwq, qaq;
      qwq = s[1][n - i] + s[0][n] - s[0][n - i], qaq = s[1][n] - s[1][n - i];
      dp[now + 1][qaq] = min(dp[now + 1][qaq], dp[now][i] + qwq);
      qwq = s[0][n - i] + s[1][n] - s[1][n - i], qaq = n - s[0][n - i];
      dp[now + 1][qaq] = min(dp[now + 1][qaq], dp[now][i] + qwq);
    }
  }
  return printf("%d\n", dp[61][0]), 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (!(c ^ 45)) f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
int n, dp[65][100005], id[100005], now, s[2][100005];
long long a[100005];
inline char cmp(int x, int y) {
  return (a[x] & ((1ll << now) - 1)) < (a[y] & ((1ll << now) - 1));
}
signed main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) a[i] = a[n] - a[i], id[i] = i;
  memset(dp, 0x3f, sizeof(dp)), dp[0][0] = 0;
  for (now = 0; now <= 60; now++) {
    sort(id + 1, id + n + 1, cmp);
    for (int i = 1; i <= n; i++)
      s[0][i] = s[0][i - 1], s[1][i] = s[1][i - 1],
      s[(a[id[i]] >> now) & 1][i]++;
    for (int i = 0; i <= n; i++) {
      int qwq, qaq;
      qwq = s[1][n - i] + s[0][n] - s[0][n - i], qaq = s[1][n] - s[1][n - i];
      dp[now + 1][qaq] = min(dp[now + 1][qaq], dp[now][i] + qwq);
      qwq = s[0][n - i] + s[1][n] - s[1][n - i], qaq = n - s[0][n - i];
      dp[now + 1][qaq] = min(dp[now + 1][qaq], dp[now][i] + qwq);
    }
  }
  return printf("%d\n", dp[61][0]), 0;
}

```