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 INF = 2e18 + 5;
int main() {
  int n, k;
  cin >> n >> k;
  map<int, int> f;
  vector<pair<long long, long long> > v;
  for (int x, i = 0; i < n; i++) {
    cin >> x;
    f[x]++;
  }
  for (auto x : f) v.emplace_back(x.first, x.second);
  n = v.size();
  vector<long long> leftSum(n);
  vector<long long> rightSum(n);
  vector<long long> amountL(n);
  vector<long long> amountR(n);
  amountR[n - 1] = v[n - 1].second;
  amountL[0] = v[0].second;
  rightSum[n - 1] = v[n - 1].first * v[n - 1].second;
  leftSum[0] = v[0].first * v[0].second;
  for (int i = 1; i < n; i++) {
    leftSum[i] = leftSum[i - 1] + v[i].first * v[i].second;
    amountL[i] = amountL[i - 1] + v[i].second;
  }
  for (int i = n - 2; i >= 0; i--) {
    rightSum[i] = rightSum[i + 1] + v[i].first * v[i].second;
    amountR[i] = amountR[i + 1] + v[i].second;
  }
  for (int i = 0; i < n; i++) {
    amountR[i] -= v[i].second;
    amountL[i] -= v[i].second;
    leftSum[i] -= v[i].first * v[i].second;
    rightSum[i] -= v[i].first * v[i].second;
  }
  long long ans = INF;
  for (int i = 0; i < n; i++) {
    long long need = k - v[i].second;
    if (need <= 0LL) {
      cout << "0\n";
      return 0;
    }
    long long costL =
        abs((v[i].first - 1) * amountL[i] - leftSum[i]) + min(amountL[i], need);
    long long costR = abs((v[i].first + 1) * amountR[i] - rightSum[i]) +
                      min(amountR[i], need);
    if (amountL[i] >= need) {
      ans = min(ans, costL);
    }
    if (amountR[i] >= need) {
      ans = min(ans, costR);
    }
    ans = min(ans, costL + abs((v[i].first + 1LL) * amountR[i] - rightSum[i]) +
                       max(need - amountL[i], 0LL));
    ans = min(ans, costR + abs((v[i].first - 1LL) * amountL[i] - leftSum[i]) +
                       max(need - amountR[i], 0LL));
  }
  cout << ans << "\n";
  return 0;
}