#include <bits/stdc++.h>
using namespace std;
const int NMAX = 100010;
const int E = 1e-8;
const int PI = acos(-1);
const int MOD = 1000000007;
const int BMAX = 60;
int n, m;
long long a[NMAX];
long long b[NMAX];
int dp[BMAX][NMAX];
int cur = 0;
int cnt0[NMAX], cnt1[NMAX];
int calcBit(long long a) {
  int tmp = 0;
  while (a > 0) {
    tmp++;
    a /= 2;
  }
  return tmp + 1;
}
bool cmp(long long x, long long y) {
  return (x & ((1LL << cur) - 1)) > (y & ((1LL << cur) - 1));
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = (1); i <= (n); i++) b[i] = a[n] - a[i];
  m = calcBit(b[1]);
  for (int i = (0); i <= (m); i++)
    for (int j = (0); j <= (n); j++) dp[i][j] = MOD;
  dp[0][0] = 0;
  for (int i = (0); i <= (m - 1); i++) {
    cur = i;
    if (i > 0) sort(b + 1, b + n + 1, cmp);
    for (int j = (1); j <= (n); j++)
      cnt1[j] = cnt1[j - 1] + (((b[j]) & (1LL << (i))) > 0);
    for (int j = (1); j <= (n); j++)
      cnt0[j] = cnt0[j - 1] + !(((b[j]) & (1LL << (i))) > 0);
    for (int j = (0); j <= (n); j++) {
      for (int t = (0); t <= (1); t++) {
        if (dp[i][j] == MOD) continue;
        int n_carry, n_ones;
        if (t == 0) {
          n_carry = cnt1[j];
          n_ones = cnt0[j] + (cnt1[n] - cnt1[j]);
        } else {
          n_carry = j + (cnt1[n] - cnt1[j]);
          n_ones = cnt1[j] + (cnt0[n] - cnt0[j]);
        }
        dp[i + 1][n_carry] = min(dp[i + 1][n_carry], dp[i][j] + n_ones);
      }
    }
  }
  cout << dp[m][0] << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int NMAX = 100010;
const int E = 1e-8;
const int PI = acos(-1);
const int MOD = 1000000007;
const int BMAX = 60;
int n, m;
long long a[NMAX];
long long b[NMAX];
int dp[BMAX][NMAX];
int cur = 0;
int cnt0[NMAX], cnt1[NMAX];
int calcBit(long long a) {
  int tmp = 0;
  while (a > 0) {
    tmp++;
    a /= 2;
  }
  return tmp + 1;
}
bool cmp(long long x, long long y) {
  return (x & ((1LL << cur) - 1)) > (y & ((1LL << cur) - 1));
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = (1); i <= (n); i++) b[i] = a[n] - a[i];
  m = calcBit(b[1]);
  for (int i = (0); i <= (m); i++)
    for (int j = (0); j <= (n); j++) dp[i][j] = MOD;
  dp[0][0] = 0;
  for (int i = (0); i <= (m - 1); i++) {
    cur = i;
    if (i > 0) sort(b + 1, b + n + 1, cmp);
    for (int j = (1); j <= (n); j++)
      cnt1[j] = cnt1[j - 1] + (((b[j]) & (1LL << (i))) > 0);
    for (int j = (1); j <= (n); j++)
      cnt0[j] = cnt0[j - 1] + !(((b[j]) & (1LL << (i))) > 0);
    for (int j = (0); j <= (n); j++) {
      for (int t = (0); t <= (1); t++) {
        if (dp[i][j] == MOD) continue;
        int n_carry, n_ones;
        if (t == 0) {
          n_carry = cnt1[j];
          n_ones = cnt0[j] + (cnt1[n] - cnt1[j]);
        } else {
          n_carry = j + (cnt1[n] - cnt1[j]);
          n_ones = cnt1[j] + (cnt0[n] - cnt0[j]);
        }
        dp[i + 1][n_carry] = min(dp[i + 1][n_carry], dp[i][j] + n_ones);
      }
    }
  }
  cout << dp[m][0] << endl;
  return 0;
}

```