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;
int A[200002];
long long PS[200001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ((void)0);
  ((void)0);
  ((void)0);
  int N, K;
  long long ans, m = 0x7fffffffffffffffLL;
  cin >> N >> K;
  for (int i = 1; i <= N; i++) cin >> A[i];
  sort(A + 1, A + N + 1);
  for (int i = 1; i <= N; i++) {
    PS[i] = PS[i - 1] + A[i];
    if (i + K - 1 <= N && A[i] == A[i + K - 1]) {
      cout << "0\n";
      return 0;
    }
  }
  for (int i = K; i <= N; i++) {
    ans = 1LL * K * A[K] - PS[K] - i + K;
    if (A[i] != A[i + 1]) break;
  }
  for (int i = N - K; i >= 0; i--) {
    ans = min(ans, PS[N] - PS[N - K] - 1LL * K * A[N - K + 1] - (N - K - i));
    if (A[i] != A[i + 1]) break;
  }
  for (int i = 1; i <= N; i++)
    m = min(m, PS[N] - 2 * PS[i] - 1LL * (N - 2 * i) * A[i]);
  cout << min(ans, m - N + K) << '\n';
  return 0;
}