Problem: You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and the required number of equal elements.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long all[200005];
bool cmp(long long a, long long b) { return a < b; }
struct A {
  long long how;
  long long con;
  long long a;
  long long b;
  long long xx;
  long long yy;
} temp[200005];
int main() {
  long long N, K, i, now = 0, ans = 1e18;
  scanf("%lld %lld", &N, &K);
  for (i = 0; i < N; i++) scanf("%lld", &all[i]);
  sort(all, all + N, cmp);
  now = -1;
  for (i = 0; i < N; i++) {
    if (i == 0 || all[i] != all[i - 1]) {
      now++;
      temp[now].how = all[i];
      temp[now].con = 0;
      temp[now].xx = 0;
      temp[now].yy = 0;
      temp[now].a = 0;
      temp[now].b = 0;
    }
    temp[now].con++;
    if (temp[now].con >= K) ans = 0;
  }
  now++;
  temp[0].a = temp[0].con;
  temp[0].xx = 0;
  for (i = 1; i < now; i++) {
    temp[i].a = temp[i - 1].a + temp[i].con;
    temp[i].xx =
        temp[i - 1].xx + temp[i - 1].a * (temp[i].how - temp[i - 1].how);
    if (temp[i].a >= K) ans = min(ans, temp[i].xx - (temp[i].a - K));
  }
  temp[now - 1].b = temp[now - 1].con;
  temp[now - 1].yy = 0;
  for (i = now - 2; i >= 0; i--) {
    temp[i].b = temp[i + 1].b + temp[i].con;
    temp[i].yy =
        temp[i + 1].yy + temp[i + 1].b * (temp[i + 1].how - temp[i].how);
    if (temp[i].b >= K) ans = min(ans, temp[i].yy - (temp[i].b - K));
  }
  for (i = 0; i < now; i++) {
    if (temp[i].a >= K)
      ans = min(ans, temp[i].xx);
    else {
      if (i + 1 != now)
        ans = min(ans, temp[i].xx + temp[i + 1].yy + (K - temp[i].a) +
                           temp[i + 1].b * (temp[i + 1].how - temp[i].how - 1));
    }
    if (temp[i].b >= K)
      ans = min(ans, temp[i].yy);
    else {
      if (i)
        ans = min(ans, temp[i].yy + temp[i - 1].xx + (K - temp[i].b) +
                           temp[i - 1].a * (temp[i].how - temp[i - 1].how - 1));
    }
  }
  printf("%lld\n", max(ans, (long long)0));
  return 0;
}