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 long long maxn = 200005;
long long tree[2][maxn];
void add(long long k, long long v, long long t) {
  for (long long i = k; i < maxn; i += i & -i) tree[t][i] += v;
}
long long sum(long long k, long long t) {
  long long res = 0;
  for (long long i = k; i >= 1; i -= i & -i) res += tree[t][i];
  return res;
}
long long a[maxn];
map<long long, long long> g;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  set<long long> s;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  vector<long long> v(s.begin(), s.end());
  for (long long i = 0; i < v.size(); i++) g[v[i]] = i + 1;
  for (long long i = 0; i < n; i++) {
    add(g[a[i]], a[i], 0);
    add(g[a[i]], 1, 1);
  }
  long long best = 1e18;
  for (long long i = 0; i < n; i++) {
    long long cnt = sum(g[a[i]], 1) - sum(g[a[i]] - 1, 1);
    if (cnt >= k) {
      best = 0;
      continue;
    }
    long long seek = k - cnt;
    long long freqLeft = sum(g[a[i]] - 1, 1);
    long long sumLeft = (a[i] - 1) * freqLeft - sum(g[a[i]] - 1, 0);
    long long freqRight = sum(maxn - 1, 1) - sum(g[a[i]], 1);
    long long sumRight =
        sum(maxn - 1, 0) - sum(g[a[i]], 0) - (a[i] + 1) * freqRight;
    if (freqLeft >= seek) best = min(best, sumLeft + seek);
    if (freqRight >= seek) best = min(best, sumRight + seek);
    best = min(best, sumLeft + sumRight + seek);
  }
  cout << best << '\n';
}