#include <bits/stdc++.h>
using namespace std;
long long n, a[100010];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
long long dp[65][100010], id[100010], now;
long long cmp(long long aa, long long bb) {
  return (a[aa] & ((1ll << now) - 1ll)) < (a[bb] & ((1ll << now) - 1ll));
}
long long sum[2][100010];
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; ++i) a[i] = a[n] - a[i], id[i] = i;
  memset(dp, 63, sizeof(dp));
  dp[0][0] = 0;
  for (long long k = 0; k <= 60; k++) {
    now = k;
    sort(id + 1, id + n + 1, cmp);
    for (long long 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 (long long i = 0; i <= n; ++i) {
      long long w, jin;
      w = sum[1][n - i] + sum[0][n] - sum[0][n - i];
      jin = sum[1][n] - sum[1][n - i];
      dp[k + 1][jin] = min(dp[k + 1][jin], dp[k][i] + w);
      w = sum[0][n - i] + sum[1][n] - sum[1][n - i];
      jin = n - sum[0][n - i];
      dp[k + 1][jin] = min(dp[k + 1][jin], dp[k][i] + w);
    }
  }
  cout << dp[61][0] << endl;
}

### Prompt

Your task is to create 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, a[100010];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
long long dp[65][100010], id[100010], now;
long long cmp(long long aa, long long bb) {
  return (a[aa] & ((1ll << now) - 1ll)) < (a[bb] & ((1ll << now) - 1ll));
}
long long sum[2][100010];
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; ++i) a[i] = a[n] - a[i], id[i] = i;
  memset(dp, 63, sizeof(dp));
  dp[0][0] = 0;
  for (long long k = 0; k <= 60; k++) {
    now = k;
    sort(id + 1, id + n + 1, cmp);
    for (long long 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 (long long i = 0; i <= n; ++i) {
      long long w, jin;
      w = sum[1][n - i] + sum[0][n] - sum[0][n - i];
      jin = sum[1][n] - sum[1][n - i];
      dp[k + 1][jin] = min(dp[k + 1][jin], dp[k][i] + w);
      w = sum[0][n - i] + sum[1][n] - sum[1][n - i];
      jin = n - sum[0][n - i];
      dp[k + 1][jin] = min(dp[k + 1][jin], dp[k][i] + w);
    }
  }
  cout << dp[61][0] << endl;
}

```