#include <bits/stdc++.h>
using namespace std;
long long n, now, a[100010], id[100010], dp[64][100010], sum[2][100010];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
bool cmp(long long x, long long y) {
  return (a[x] & ((1ll << now) - 1)) < (a[y] & ((1ll << now) - 1));
}
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, 0x3f3f3f3f3f3f, sizeof(dp));
  dp[0][0] = 0;
  for (long long i = 0; i <= 60; i++) {
    now = i;
    sort(id + 1, id + n + 1, cmp);
    for (long long j = 1; j <= n; j++) {
      sum[0][j] = sum[0][j - 1];
      sum[1][j] = sum[1][j - 1];
      sum[(a[id[j]] >> i) & 1][j]++;
    }
    for (long long j = 0; j <= n; j++) {
      dp[i + 1][sum[1][n] - sum[1][n - j]] =
          min(dp[i + 1][sum[1][n] - sum[1][n - j]],
              dp[i][j] + sum[1][n - j] + sum[0][n] - sum[0][n - j]);
      dp[i + 1][n - sum[0][n - j]] =
          min(dp[i + 1][n - sum[0][n - j]],
              dp[i][j] + sum[0][n - j] + sum[1][n] - sum[1][n - j]);
    }
  }
  printf("%lld\n", dp[61][0]);
  return 0;
}

### Prompt

Please create a solution in cpp 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, now, a[100010], id[100010], dp[64][100010], sum[2][100010];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
bool cmp(long long x, long long y) {
  return (a[x] & ((1ll << now) - 1)) < (a[y] & ((1ll << now) - 1));
}
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, 0x3f3f3f3f3f3f, sizeof(dp));
  dp[0][0] = 0;
  for (long long i = 0; i <= 60; i++) {
    now = i;
    sort(id + 1, id + n + 1, cmp);
    for (long long j = 1; j <= n; j++) {
      sum[0][j] = sum[0][j - 1];
      sum[1][j] = sum[1][j - 1];
      sum[(a[id[j]] >> i) & 1][j]++;
    }
    for (long long j = 0; j <= n; j++) {
      dp[i + 1][sum[1][n] - sum[1][n - j]] =
          min(dp[i + 1][sum[1][n] - sum[1][n - j]],
              dp[i][j] + sum[1][n - j] + sum[0][n] - sum[0][n - j]);
      dp[i + 1][n - sum[0][n - j]] =
          min(dp[i + 1][n - sum[0][n - j]],
              dp[i][j] + sum[0][n - j] + sum[1][n] - sum[1][n - j]);
    }
  }
  printf("%lld\n", dp[61][0]);
  return 0;
}

```