#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const long long MAXN = 1e5 + 11;
long long N, A[MAXN], f[62][MAXN], id[MAXN];
long long now, Sum[2][MAXN];
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++) id[i] = i, A[i] = A[N] - A[i];
  memset(f, 127 / 3, sizeof(f));
  f[0][0] = 0;
  for (long long i = 0; i < 61; i++) {
    now = i;
    sort(id + 1, id + N + 1, cmp);
    memset(Sum, 0, sizeof(Sum));
    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++) {
      long long W = Sum[1][N - j] + Sum[0][N] - Sum[0][N - j],
                nex = Sum[1][N] - Sum[1][N - j];
      f[i + 1][nex] = min(f[i + 1][nex], f[i][j] + W);
      W = Sum[0][N - j] + Sum[1][N] - Sum[1][N - j], nex = N - Sum[0][N - j];
      f[i + 1][nex] = min(f[i + 1][nex], f[i][j] + W);
    }
  }
  printf("%lld\n", f[61][0]);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
inline long long read() {
  long long f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const long long MAXN = 1e5 + 11;
long long N, A[MAXN], f[62][MAXN], id[MAXN];
long long now, Sum[2][MAXN];
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++) id[i] = i, A[i] = A[N] - A[i];
  memset(f, 127 / 3, sizeof(f));
  f[0][0] = 0;
  for (long long i = 0; i < 61; i++) {
    now = i;
    sort(id + 1, id + N + 1, cmp);
    memset(Sum, 0, sizeof(Sum));
    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++) {
      long long W = Sum[1][N - j] + Sum[0][N] - Sum[0][N - j],
                nex = Sum[1][N] - Sum[1][N - j];
      f[i + 1][nex] = min(f[i + 1][nex], f[i][j] + W);
      W = Sum[0][N - j] + Sum[1][N] - Sum[1][N - j], nex = N - Sum[0][N - j];
      f[i + 1][nex] = min(f[i + 1][nex], f[i][j] + W);
    }
  }
  printf("%lld\n", f[61][0]);
  return 0;
}

```