#include <bits/stdc++.h>
using namespace std;
long long ara[100005];
int carry[62][100005][2];
int ans[62][100005][2];
int dp[62][100005];
int bit;
int n;
bool cmp(long long a, long long b) {
  return a % (1LL << bit) > b % (1LL << bit);
}
int cnt(int k, int cr) {
  if (k == 62) {
    return cr;
  }
  if (dp[k][cr] != -1) return dp[k][cr];
  int x = ans[k][cr][0] + cnt(k + 1, carry[k][cr][0]);
  int y = ans[k][cr][1] + cnt(k + 1, carry[k][cr][1]);
  return dp[k][cr] = min(x, y);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", ara + i);
  sort(ara + 1, ara + n + 1);
  for (int i = 1; i <= n; i++) ara[i] = ara[n] - ara[i];
  for (int i = 1; i <= n; i++) {
    if (ara[i] & 1)
      ans[0][0][0]++, carry[0][0][1]++;
    else
      ans[0][0][1]++;
  }
  for (int k = 1; k < 62; k++) {
    bit = k;
    sort(ara + 1, ara + n + 1, cmp);
    for (int i = 1; i <= n; i++) {
      if (ara[i] & (1LL << k)) {
        carry[k][i][0]++;
        carry[k][0][1]++;
        ans[k][0][0]++, ans[k][i][0]--;
        ans[k][i][1]++;
      } else {
        carry[k][i][1]++;
        ans[k][i][0]++;
        ans[k][0][1]++, ans[k][i][1]--;
      }
    }
    for (int i = 1; i <= n; i++) {
      carry[k][i][0] += carry[k][i - 1][0];
      carry[k][i][1] += carry[k][i - 1][1];
      ans[k][i][0] += ans[k][i - 1][0];
      ans[k][i][1] += ans[k][i - 1][1];
    }
  }
  memset(dp, -1, sizeof(dp));
  printf("%d\n", cnt(0, 0));
  return 0;
}

### Prompt

Please formulate 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 ara[100005];
int carry[62][100005][2];
int ans[62][100005][2];
int dp[62][100005];
int bit;
int n;
bool cmp(long long a, long long b) {
  return a % (1LL << bit) > b % (1LL << bit);
}
int cnt(int k, int cr) {
  if (k == 62) {
    return cr;
  }
  if (dp[k][cr] != -1) return dp[k][cr];
  int x = ans[k][cr][0] + cnt(k + 1, carry[k][cr][0]);
  int y = ans[k][cr][1] + cnt(k + 1, carry[k][cr][1]);
  return dp[k][cr] = min(x, y);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", ara + i);
  sort(ara + 1, ara + n + 1);
  for (int i = 1; i <= n; i++) ara[i] = ara[n] - ara[i];
  for (int i = 1; i <= n; i++) {
    if (ara[i] & 1)
      ans[0][0][0]++, carry[0][0][1]++;
    else
      ans[0][0][1]++;
  }
  for (int k = 1; k < 62; k++) {
    bit = k;
    sort(ara + 1, ara + n + 1, cmp);
    for (int i = 1; i <= n; i++) {
      if (ara[i] & (1LL << k)) {
        carry[k][i][0]++;
        carry[k][0][1]++;
        ans[k][0][0]++, ans[k][i][0]--;
        ans[k][i][1]++;
      } else {
        carry[k][i][1]++;
        ans[k][i][0]++;
        ans[k][0][1]++, ans[k][i][1]--;
      }
    }
    for (int i = 1; i <= n; i++) {
      carry[k][i][0] += carry[k][i - 1][0];
      carry[k][i][1] += carry[k][i - 1][1];
      ans[k][i][0] += ans[k][i - 1][0];
      ans[k][i][1] += ans[k][i - 1][1];
    }
  }
  memset(dp, -1, sizeof(dp));
  printf("%d\n", cnt(0, 0));
  return 0;
}

```