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;
int n, k, a[200005], scnt;
long long s[200005], ans = ~(1LL << 63);
map<int, int> mp;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    mp[a[i]]++;
    s[i] = s[i - 1] + a[i];
  }
  for (auto i : mp) {
    int x = i.first, cnt = i.second;
    if (cnt >= k) return puts("0"), 0;
    int l = k - cnt;
    long long f1 = 1LL * (x - 1) * scnt - s[scnt];
    long long f2 = (s[n] - s[scnt + cnt]) - 1LL * (x + 1) * (n - scnt - cnt);
    if (scnt >= l) ans = min(ans, f1 + l);
    if (n - scnt - cnt >= l) ans = min(ans, f2 + l);
    ans = min(ans, l + f1 + f2);
    scnt += cnt;
  }
  printf("%lld\n", ans);
  return 0;
}