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 MX = 200005;
long long n, k;
long long a[MX];
long long sa[MX];
long long querySum(long long b, long long e) {
  if (b > e) return 0;
  long long s0 = sa[e];
  if (b > 0) s0 -= sa[b - 1];
  return s0;
}
long long solve() {
  memset(sa, 0, sizeof(sa));
  for (long long i = 0; i < n; i++) {
    sa[i] += a[i];
    if (i > 0) sa[i] += sa[i - 1];
  }
  long long res = sa[n - 1];
  long long b = 0, e = 0;
  while (b < n && e < n) {
    while (e < n && a[e] == a[b]) e++;
    long long r = e - b;
    if (r >= k) return 0;
    long long t = a[b];
    long long leftNum = 0, leftCnt = 0;
    long long rightNum = 0, rightCnt = 0;
    if (b - 1 >= 0) {
      leftNum = (t - 1) * b - querySum(0, b - 1);
      leftCnt = b;
    }
    if (e <= n - 1) {
      rightCnt = n - e;
      rightNum = -(t + 1) * rightCnt + querySum(e, n - 1);
    }
    if (leftCnt + r >= k) {
      res = min(res, leftNum + k - r);
    }
    if (rightCnt + r >= k) {
      res = min(res, rightNum + k - r);
    }
    res = min(res, leftNum + rightNum + k - r);
    b = e;
  }
  return res;
}
int main() {
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  cout << solve() << endl;
  return 0;
}