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 = 0x3f3f3f3f;
const int negative_infinite = 0xcfcfcfcf;
const double pi = acos(-1);
const int mod = 1e9 + 7;
const double eps = 1e-8;
const int MAXN = 1e6 + 117;
const int MAXM = 2e5 + 117;
int n, k;
long long a[MAXN];
long long pre[MAXN], suf[MAXN];
map<int, int> l, r;
void build() {
  for (int i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + a[i];
    if (l[a[i]] == 0) l[a[i]] = i;
  }
  for (int i = n; i > 0; i--) {
    suf[i] = suf[i + 1] + a[i];
    if (r[a[i]] == 0) r[a[i]] = i;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + n + 1);
  build();
  long long ans = 1e15;
  for (int i = 1; i <= n; i++) {
    int dx = k - (r[a[i]] - l[a[i]] + 1);
    if (dx <= 0) {
      ans = 0;
      break;
    }
    long long lcost = (l[a[i]] - 1) * (a[i] - 1) - pre[l[a[i]] - 1];
    long long rcost = suf[r[a[i]] + 1] - (n - r[a[i]]) * (a[i] + 1);
    if (i >= k) ans = min(ans, lcost + dx);
    if (n - i + 1 >= k) ans = min(ans, rcost + dx);
    ans = min(ans, lcost + rcost + dx);
  }
  printf("%lld\n", ans);
  return 0;
}