#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
int n, m;
long long a[100005];
int sum[100005], sa[100005];
int f[65][100005];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
long long ta[100005], tb[100005];
void my_sort(long long *a, int x) {
  int na = 0, nb = 0, pps = 0;
  for (int i = 1; i <= n; i++)
    if ((a[i] >> x) & 1)
      ta[++na] = a[i];
    else
      tb[++nb] = a[i];
  for (int i = 1; i <= nb; i++) a[++pps] = tb[i];
  for (int i = 1; i <= na; i++) a[++pps] = ta[i];
  for (int i = n; i; i--) sa[i] = sa[i + 1] + ((a[i] >> (x + 1)) & 1);
}
int bit_count(long long x) { return x ? __builtin_clzll(x) : 64; }
int main() {
  n = read();
  int mx = 0;
  long long fuckpps = 0;
  for (int i = 1; i <= n; i++) a[i] = read(), fuckpps = max(fuckpps, a[i]);
  for (int i = 1; i <= n; i++) {
    a[i] = fuckpps - a[i], mx = max(mx, 64 - bit_count(a[i]));
    for (int j = 1; j <= 60; j++) sum[j] += (a[i] >> (j - 1)) & 1;
  }
  memset(f, 63, sizeof(f));
  f[0][n + 1] = 0;
  for (int i = n; i; i--) sa[i] = sa[i + 1] + (a[i] & 1);
  for (int i = 0; i < mx; i++) {
    for (int j = 1; j <= n + 1; j++)
      if (f[i][j] < inf) {
        f[i + 1][n - sa[j] + 1] =
            min(f[i + 1][n - sa[j] + 1],
                f[i][j] + (sa[1] - sa[j]) + (n - j + 1 - sa[j]));
        f[i + 1][j - sa[1] + sa[j]] = min(f[i + 1][j - sa[1] + sa[j]],
                                          f[i][j] + j - 1 + sa[j] * 2 - sa[1]);
      }
    my_sort(a, i);
  }
  int ans = inf;
  for (int i = 1; i <= n + 1; i++) ans = min(ans, f[mx][i] + n - i + 1);
  printf("%d\n", ans);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 inf = 1e9;
int n, m;
long long a[100005];
int sum[100005], sa[100005];
int f[65][100005];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
long long ta[100005], tb[100005];
void my_sort(long long *a, int x) {
  int na = 0, nb = 0, pps = 0;
  for (int i = 1; i <= n; i++)
    if ((a[i] >> x) & 1)
      ta[++na] = a[i];
    else
      tb[++nb] = a[i];
  for (int i = 1; i <= nb; i++) a[++pps] = tb[i];
  for (int i = 1; i <= na; i++) a[++pps] = ta[i];
  for (int i = n; i; i--) sa[i] = sa[i + 1] + ((a[i] >> (x + 1)) & 1);
}
int bit_count(long long x) { return x ? __builtin_clzll(x) : 64; }
int main() {
  n = read();
  int mx = 0;
  long long fuckpps = 0;
  for (int i = 1; i <= n; i++) a[i] = read(), fuckpps = max(fuckpps, a[i]);
  for (int i = 1; i <= n; i++) {
    a[i] = fuckpps - a[i], mx = max(mx, 64 - bit_count(a[i]));
    for (int j = 1; j <= 60; j++) sum[j] += (a[i] >> (j - 1)) & 1;
  }
  memset(f, 63, sizeof(f));
  f[0][n + 1] = 0;
  for (int i = n; i; i--) sa[i] = sa[i + 1] + (a[i] & 1);
  for (int i = 0; i < mx; i++) {
    for (int j = 1; j <= n + 1; j++)
      if (f[i][j] < inf) {
        f[i + 1][n - sa[j] + 1] =
            min(f[i + 1][n - sa[j] + 1],
                f[i][j] + (sa[1] - sa[j]) + (n - j + 1 - sa[j]));
        f[i + 1][j - sa[1] + sa[j]] = min(f[i + 1][j - sa[1] + sa[j]],
                                          f[i][j] + j - 1 + sa[j] * 2 - sa[1]);
      }
    my_sort(a, i);
  }
  int ans = inf;
  for (int i = 1; i <= n + 1; i++) ans = min(ans, f[mx][i] + n - i + 1);
  printf("%d\n", ans);
  return 0;
}

```