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>
int inp() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum;
}
int cnt[200010];
std::priority_queue<int> q[200010];
int sum[200010];
int lg[200010], k, ans = 2147483647;
void dfs(int cur) {
  lg[cur] = lg[cur >> 1] + 1;
  if ((cur << 1) <= 200000) {
    dfs(cur << 1);
    while (!q[cur << 1].empty()) {
      sum[cur] += q[cur << 1].top();
      q[cur].push(q[cur << 1].top());
      q[cur << 1].pop();
    }
  }
  if ((cur << 1 | 1) <= 200000) {
    dfs(cur << 1 | 1);
    while (!q[cur << 1 | 1].empty()) {
      sum[cur] += q[cur << 1 | 1].top();
      q[cur].push(q[cur << 1 | 1].top());
      q[cur << 1 | 1].pop();
    }
  }
  for (int i = 1; i <= cnt[cur]; i++) {
    q[cur].push(lg[cur]);
    sum[cur] += lg[cur];
  }
  while (q[cur].size() > k) {
    sum[cur] -= q[cur].top();
    q[cur].pop();
  }
  if (q[cur].size() >= k) {
    ans = std::min(ans, sum[cur] - (lg[cur] * k));
  }
}
int main() {
  int n = inp();
  k = inp();
  for (int i = 1; i <= n; i++) cnt[inp()]++;
  lg[0] = -1;
  dfs(1);
  printf("%d\n", ans);
}