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 INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};
const int N = 2e5 + 2;
int n, k, a[N];
long long psum[N], ssum[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) psum[i] = psum[i - 1] + a[i];
  for (int i = n; i >= 1; i--) ssum[i] = ssum[i + 1] + a[i];
  long long res = 1ll * INF * INF;
  for (int i = 1; i <= n; i++) {
    if (a[i] == a[i - 1]) continue;
    int j = i;
    while (j < n && a[j + 1] == a[j]) j++;
    int x = i - 1;
    int y = n - j;
    int need = k - (j - i + 1);
    if (need <= 0) return cout << 0, 0;
    long long costL = 1ll * a[i] * x - psum[i - 1];
    long long costR = ssum[j + 1] - 1ll * a[i] * y;
    res = min(res, costL + costR - (x + y - need));
    if (x >= need) res = min(res, costL - (x - need));
    if (y >= need) res = min(res, costR - (y - need));
  }
  cout << res;
  return 0;
}