#include <bits/stdc++.h>
const int maxn = 1e5 + 5, lim = 62, maxm = 65;
const long long inf = 1e18;
template <typename T>
inline T min(T a, T b) {
  return a < b ? a : b;
}
int n;
long long a[maxn], mod, dp[maxm][maxn];
inline bool Comp(long long a, long long b) { return (a % mod) > (b % mod); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  mod = 1ll << lim;
  std::sort(a + 1, a + n + 1, Comp);
  for (int i = n; i >= 1; i--) a[i] = a[1] - a[i];
  for (int i = 0; i <= lim; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = inf;
  dp[0][0] = 0;
  for (int i = 0; i < lim; i++) {
    mod = 1ll << i;
    std::sort(a + 1, a + n + 1, Comp);
    long long cnt = 0, tot = 0;
    for (int j = 1; j <= n; j++) {
      cnt += a[j] & mod ? 1 : 0;
    }
    for (int j = 0; j <= n; j++) {
      tot += a[j] & mod ? 1 : 0;
      dp[i + 1][cnt - tot + j] =
          min(dp[i + 1][cnt - tot + j], dp[i][j] + tot + (n - j) - (cnt - tot));
      dp[i + 1][tot] = min(dp[i + 1][tot], dp[i][j] + j - tot + cnt - tot);
    }
  }
  printf("%lld\n", dp[lim][0]);
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
const int maxn = 1e5 + 5, lim = 62, maxm = 65;
const long long inf = 1e18;
template <typename T>
inline T min(T a, T b) {
  return a < b ? a : b;
}
int n;
long long a[maxn], mod, dp[maxm][maxn];
inline bool Comp(long long a, long long b) { return (a % mod) > (b % mod); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  mod = 1ll << lim;
  std::sort(a + 1, a + n + 1, Comp);
  for (int i = n; i >= 1; i--) a[i] = a[1] - a[i];
  for (int i = 0; i <= lim; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = inf;
  dp[0][0] = 0;
  for (int i = 0; i < lim; i++) {
    mod = 1ll << i;
    std::sort(a + 1, a + n + 1, Comp);
    long long cnt = 0, tot = 0;
    for (int j = 1; j <= n; j++) {
      cnt += a[j] & mod ? 1 : 0;
    }
    for (int j = 0; j <= n; j++) {
      tot += a[j] & mod ? 1 : 0;
      dp[i + 1][cnt - tot + j] =
          min(dp[i + 1][cnt - tot + j], dp[i][j] + tot + (n - j) - (cnt - tot));
      dp[i + 1][tot] = min(dp[i + 1][tot], dp[i][j] + j - tot + cnt - tot);
    }
  }
  printf("%lld\n", dp[lim][0]);
  return 0;
}

```