#include <bits/stdc++.h>
using namespace std;
int dp[2][100010], bit, n;
long long a[100010];
inline int comp(long long x, long long y) {
  return (x % (1ll << (bit + 1))) > (y % (1ll << (bit + 1)));
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  int cur = 0;
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < 63; i++) {
    cur ^= 1;
    int cnt1 = 0, cnt2 = 0;
    memset(dp[cur], 0x3f, sizeof(dp[cur]));
    for (int j = 0; j < n; j++) {
      if (a[j] & (1ll << i)) cnt2++;
    }
    for (int j = 0; j <= n; j++) {
      if (dp[cur ^ 1][j] != 0x3f3f3f3f) {
        dp[cur][j + cnt2] =
            min(dp[cur][j + cnt2], dp[cur ^ 1][j] + j - cnt1 + cnt2);
        dp[cur][cnt1] =
            min(dp[cur][cnt1], dp[cur ^ 1][j] + cnt1 + (n - j - cnt2));
      }
      if (a[j] & (1ll << i)) {
        cnt1++;
        cnt2--;
      }
    }
    if ((1ll << i) > mx) break;
    bit = i;
    sort(a, a + n, comp);
  }
  printf("%d\n", dp[cur][0]);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int dp[2][100010], bit, n;
long long a[100010];
inline int comp(long long x, long long y) {
  return (x % (1ll << (bit + 1))) > (y % (1ll << (bit + 1)));
}
int main() {
  long long mx = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    mx = max(mx, a[i]);
  }
  int cur = 0;
  memset(dp, 0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < 63; i++) {
    cur ^= 1;
    int cnt1 = 0, cnt2 = 0;
    memset(dp[cur], 0x3f, sizeof(dp[cur]));
    for (int j = 0; j < n; j++) {
      if (a[j] & (1ll << i)) cnt2++;
    }
    for (int j = 0; j <= n; j++) {
      if (dp[cur ^ 1][j] != 0x3f3f3f3f) {
        dp[cur][j + cnt2] =
            min(dp[cur][j + cnt2], dp[cur ^ 1][j] + j - cnt1 + cnt2);
        dp[cur][cnt1] =
            min(dp[cur][cnt1], dp[cur ^ 1][j] + cnt1 + (n - j - cnt2));
      }
      if (a[j] & (1ll << i)) {
        cnt1++;
        cnt2--;
      }
    }
    if ((1ll << i) > mx) break;
    bit = i;
    sort(a, a + n, comp);
  }
  printf("%d\n", dp[cur][0]);
  return 0;
}

```