#include <bits/stdc++.h>
using namespace std;
long long n, sum[2][100005], id[100005], a[100005], nowk, dp[65][100006], mx,
    M = 60, inf = 1e18;
long long cmp(long long x, long long y) {
  long long qwq = (1ll << nowk) - 1;
  return (a[x] & qwq) < (a[y] & qwq);
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    id[i] = i;
    mx = max(mx, a[i]);
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = mx - a[i];
  }
  for (long long i = 0; i <= M + 1; i++) {
    for (long long t = 0; t <= n; t++) {
      dp[i][t] = inf;
    }
  }
  dp[0][0] = 0;
  for (long long i = 0; i <= M; i++) {
    nowk = i;
    sort(id + 1, id + n + 1, cmp);
    for (long long t = 1; t <= n; t++)
      sum[0][t] = sum[0][t - 1], sum[1][t] = sum[1][t - 1],
      sum[(a[id[t]] >> i) & 1][t]++;
    for (long long t = 0; t <= n; t++) {
      long long nowv, nex;
      nowv = sum[1][n - t] + sum[0][n] - sum[0][n - t],
      nex = sum[1][n] - sum[1][n - t];
      dp[i + 1][nex] = min(dp[i + 1][nex], dp[i][t] + nowv);
      nowv = sum[0][n - t] + sum[1][n] - sum[1][n - t], nex = n - sum[0][n - t];
      dp[i + 1][nex] = min(dp[i + 1][nex], dp[i][t] + nowv);
    }
  }
  printf("%lld", dp[61][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;
long long n, sum[2][100005], id[100005], a[100005], nowk, dp[65][100006], mx,
    M = 60, inf = 1e18;
long long cmp(long long x, long long y) {
  long long qwq = (1ll << nowk) - 1;
  return (a[x] & qwq) < (a[y] & qwq);
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    id[i] = i;
    mx = max(mx, a[i]);
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = mx - a[i];
  }
  for (long long i = 0; i <= M + 1; i++) {
    for (long long t = 0; t <= n; t++) {
      dp[i][t] = inf;
    }
  }
  dp[0][0] = 0;
  for (long long i = 0; i <= M; i++) {
    nowk = i;
    sort(id + 1, id + n + 1, cmp);
    for (long long t = 1; t <= n; t++)
      sum[0][t] = sum[0][t - 1], sum[1][t] = sum[1][t - 1],
      sum[(a[id[t]] >> i) & 1][t]++;
    for (long long t = 0; t <= n; t++) {
      long long nowv, nex;
      nowv = sum[1][n - t] + sum[0][n] - sum[0][n - t],
      nex = sum[1][n] - sum[1][n - t];
      dp[i + 1][nex] = min(dp[i + 1][nex], dp[i][t] + nowv);
      nowv = sum[0][n - t] + sum[1][n] - sum[1][n - t], nex = n - sum[0][n - t];
      dp[i + 1][nex] = min(dp[i + 1][nex], dp[i][t] + nowv);
    }
  }
  printf("%lld", dp[61][0]);
}

```