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;
multiset<int> s[200005];
int num[200005];
int sz[200005];
long long sum[200005];
int main() {
  int n, k, t;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t);
    int cost = 0;
    while (true) {
      s[t].insert(cost);
      sz[t]++;
      sum[t] += cost;
      if (sz[t] > k) {
        sz[t]--;
        auto it = s[t].end();
        it--;
        sum[t] -= *it;
        s[t].erase(it);
      }
      if (t == 0) break;
      cost++;
      t /= 2;
    }
  }
  long long ans = 1e18;
  for (int i = 0; i <= 200000; i++) {
    if (sz[i] == k) {
      ans = min(ans, sum[i]);
    }
  }
  printf("%lld\n", ans);
}