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 maxn = 2e5 + 1;
map<long long, long long> p;
vector<pair<long long, long long>> v;
long long k, sum1[maxn], sum2[maxn], num1[maxn], num2[maxn], a[maxn];
int main() {
  int n;
  scanf("%d %lld", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]), p[a[i]]++;
  for (auto it = p.begin(); it != p.end(); it++)
    v.push_back({it->first, it->second});
  long long t = 0, tt = 0, ans = 1e18;
  for (int i = 0; i < v.size(); ++i) {
    num1[i] = t;
    sum1[i] = tt;
    t += v[i].second;
    tt += v[i].first * v[i].second;
  }
  t = tt = 0;
  for (int i = v.size() - 1; i >= 0; --i) {
    num2[i] = t;
    sum2[i] = tt;
    t += v[i].second;
    tt += v[i].first * v[i].second;
  }
  for (int i = 0; i < v.size(); ++i) {
    if (v[i].second >= k) {
      puts("0");
      return 0;
    }
    long long left = num1[i] * (v[i].first - 1) - sum1[i],
              right = sum2[i] - num2[i] * (v[i].first + 1);
    if (v[i].second + num1[i] >= k) ans = min(ans, left + k - v[i].second);
    if (v[i].second + num2[i] >= k) ans = min(ans, right + k - v[i].second);
    ans = min(ans, left + right + k - v[i].second);
  }
  printf("%lld\n", ans);
}