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;
const int MAX = 2 * 1e5 + 5;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int N, K, A[MAX], table[20][MAX];
int main() {
  cin >> N >> K;
  for (int i = 0; i < (N); ++i) cin >> A[i];
  for (int i = 0; i < (N); ++i) table[0][i] = A[i];
  vector<int> b(N, 0);
  for (int i = 0; i < (N); ++i) {
    for (int j = 0; j < (19); ++j) {
      table[j + 1][i] = table[j][i] / 2;
      b[i]++;
      if (table[j + 1][i] == 0) break;
    }
  }
  map<int, int> m, cost;
  int ans = INF;
  for (int i = 0; i < (20); ++i) {
    for (int j = 0; j < (N); ++j) {
      if (b[j] < i) continue;
      int num = table[i][j];
      m[num]++;
      cost[num] += i;
      if (m[num] >= K) ans = min(ans, cost[num]);
    }
  }
  cout << ans << endl;
}