#include <bits/stdc++.h>
using namespace std;
int N;
long long A[100005];
int F[62][100005];
int id[100005];
int nowk;
bool cmp(int a, int b) {
  return (A[a] & ((1LL << nowk) - 1)) < (A[b] & ((1LL << nowk) - 1));
}
int sum[2][100005];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
  sort(A + 1, A + N + 1);
  for (int i = 1; i <= N; i++) A[i] = A[N] - A[i], id[i] = i;
  memset(F, 63, sizeof(F));
  F[0][0] = 0;
  for (int k = 0; k <= 60; k++) {
    nowk = k;
    sort(id + 1, id + N + 1, cmp);
    for (int i = 1; i <= N; i++) {
      sum[0][i] = sum[0][i - 1], sum[1][i] = sum[1][i - 1];
      sum[(A[id[i]] >> k) & 1][i]++;
    }
    for (int i = 0; i <= N; i++) {
      int w, ni;
      w = sum[1][N - i] + sum[0][N] - sum[0][N - i],
      ni = sum[1][N] - sum[1][N - i];
      F[k + 1][ni] = min(F[k + 1][ni], F[k][i] + w);
      w = sum[0][N - i] + sum[1][N] - sum[1][N - i], ni = N - sum[0][N - i];
      F[k + 1][ni] = min(F[k + 1][ni], F[k][i] + w);
    }
  }
  printf("%d\n", F[61][0]);
}

### Prompt

Your challenge is to write 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;
int N;
long long A[100005];
int F[62][100005];
int id[100005];
int nowk;
bool cmp(int a, int b) {
  return (A[a] & ((1LL << nowk) - 1)) < (A[b] & ((1LL << nowk) - 1));
}
int sum[2][100005];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
  sort(A + 1, A + N + 1);
  for (int i = 1; i <= N; i++) A[i] = A[N] - A[i], id[i] = i;
  memset(F, 63, sizeof(F));
  F[0][0] = 0;
  for (int k = 0; k <= 60; k++) {
    nowk = k;
    sort(id + 1, id + N + 1, cmp);
    for (int i = 1; i <= N; i++) {
      sum[0][i] = sum[0][i - 1], sum[1][i] = sum[1][i - 1];
      sum[(A[id[i]] >> k) & 1][i]++;
    }
    for (int i = 0; i <= N; i++) {
      int w, ni;
      w = sum[1][N - i] + sum[0][N] - sum[0][N - i],
      ni = sum[1][N] - sum[1][N - i];
      F[k + 1][ni] = min(F[k + 1][ni], F[k][i] + w);
      w = sum[0][N - i] + sum[1][N] - sum[1][N - i], ni = N - sum[0][N - i];
      F[k + 1][ni] = min(F[k + 1][ni], F[k][i] + w);
    }
  }
  printf("%d\n", F[61][0]);
}

```