#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T &x) {
  int ch = 0, f = 0;
  x = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = 1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  if (f) x = -x;
}
const int N = 100005;
long long a[N], b[N];
unsigned long long mod;
int dp[65][N], ans, n;
inline bool cmp(long long x, long long y) { return (x % mod) > (y % mod); }
inline void chkmin(int &x, int y) { x = min(x, y); }
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  mod = 1ll << 62ll;
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; i++) b[i] = a[1] - a[i];
  memset(dp, 127, sizeof(dp)), dp[0][0] = 0;
  for (long long i = 0; i < 64; i++) {
    mod = 1ll << i;
    sort(b + 1, b + n + 1, cmp);
    int s0 = 0, s1 = 0, tot = 0, tot1 = 0;
    for (int j = 1; j <= n; j++) {
      s0 += ((1ll << i) & b[j]) ? 1 : 0;
      s1 += ((1ll << i) & b[j]) ? 0 : 1;
    }
    tot1 = s0;
    for (int j = 0; j <= n; j++) {
      if (j > 0) {
        tot += ((1ll << i) & b[j]) ? 1 : 0;
        s0 += ((1ll << i) & b[j]) ? -1 : 1;
        s1 += ((1ll << i) & b[j]) ? 1 : -1;
      }
      if (dp[i][j] == (0x7f7f7f7f)) continue;
      chkmin(dp[i + 1][tot], dp[i][j] + s0);
      chkmin(dp[i + 1][j - tot + tot1], dp[i][j] + s1);
    }
  }
  cout << dp[64][0] << endl;
  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;
template <class T>
inline void read(T &x) {
  int ch = 0, f = 0;
  x = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = 1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  if (f) x = -x;
}
const int N = 100005;
long long a[N], b[N];
unsigned long long mod;
int dp[65][N], ans, n;
inline bool cmp(long long x, long long y) { return (x % mod) > (y % mod); }
inline void chkmin(int &x, int y) { x = min(x, y); }
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  mod = 1ll << 62ll;
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; i++) b[i] = a[1] - a[i];
  memset(dp, 127, sizeof(dp)), dp[0][0] = 0;
  for (long long i = 0; i < 64; i++) {
    mod = 1ll << i;
    sort(b + 1, b + n + 1, cmp);
    int s0 = 0, s1 = 0, tot = 0, tot1 = 0;
    for (int j = 1; j <= n; j++) {
      s0 += ((1ll << i) & b[j]) ? 1 : 0;
      s1 += ((1ll << i) & b[j]) ? 0 : 1;
    }
    tot1 = s0;
    for (int j = 0; j <= n; j++) {
      if (j > 0) {
        tot += ((1ll << i) & b[j]) ? 1 : 0;
        s0 += ((1ll << i) & b[j]) ? -1 : 1;
        s1 += ((1ll << i) & b[j]) ? 1 : -1;
      }
      if (dp[i][j] == (0x7f7f7f7f)) continue;
      chkmin(dp[i + 1][tot], dp[i][j] + s0);
      chkmin(dp[i + 1][j - tot + tot1], dp[i][j] + s1);
    }
  }
  cout << dp[64][0] << endl;
  return 0;
}

```