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
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
const int mod = 998244353;
int n;
long long a[N], b[N], c[N], k, ans = 2e15;
inline long long qpow(long long a, long long b, long long m) {
  long long ans = 1ll;
  for (; b; b >>= 1, a = a * a % m)
    if (b & 1) ans = ans * a % m;
  return ans;
}
inline int read() {
  int xx = 0, ff = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = xx * 10 + ch - '0';
    ch = getchar();
  }
  return xx * ff;
}
inline int gcd(int a, int b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
inline int lcm(int a, int b) { return a / gcd(a, b) * b; }
inline void pre() {
  for (int i = 1; i <= n; i++) b[i] = b[i - 1] + a[i];
}
inline void suf() {
  for (int i = n; i >= 1; i--) c[i] = c[i + 1] + a[i];
}
inline long long minLL(long long x, long long y) { return x < y ? x : y; }
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  pre();
  suf();
  for (int l = 1, r; l <= n; l = r + 1) {
    for (r = l; a[r + 1] == a[l]; r++)
      ;
    if (r - l + 1 >= k) return cout << 0, 0;
    long long left = 1ll * (a[l] - 1) * (l - 1) - b[l - 1],
              right = c[r + 1] - 1ll * (a[r] + 1) * (n - r),
              mid = k - r + l - 1;
    if (r >= k && left + mid < ans) ans = left + mid;
    if (n - l + 1 >= k) ans = min(ans, right + mid);
    ans = min(ans, left + right + mid);
  }
  return cout << ans << endl, 0;
}
