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;
using ll = long long;
using PII = pair<int, int>;
const int mod = 998244353;
const int inf = 1 << 30;
const int maxn = 200000 + 5;
int n, k, a[maxn];
ll pre[maxn], suf[maxn];
int main() {
  scanf("%d%d", &n, &k);
  map<int, int> cnt;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    cnt[a[i]]++;
  }
  for (auto& x : cnt) {
    if (x.second >= k) {
      puts("0");
      return 0;
    }
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) {
    pre[i] = a[i] + pre[i - 1];
  }
  for (int i = n; i >= 1; i--) {
    suf[i] = suf[i + 1] + a[i];
  }
  auto qL = [&](int x) {
    int k = upper_bound(a + 1, a + 1 + n, x) - a - 1;
    return make_pair(k, 1ll * k * x - pre[k]);
  };
  auto qR = [&](int x) {
    int k = lower_bound(a + 1, a + 1 + n, x) - a;
    return make_pair(n - k + 1, suf[k] - 1ll * x * (n - k + 1));
  };
  ll ans = 1e18;
  for (auto& x : cnt) {
    auto a = qL(x.first - 1);
    auto b = qR(x.first + 1);
    if (a.first + x.second >= k) {
      ans = min(ans, a.second + k - x.second);
    }
    if (b.first + x.second >= k) {
      ans = min(ans, b.second + k - x.second);
    }
    ans = min(ans, a.second + b.second + k - x.second);
  }
  printf("%I64d\n", ans);
  return 0;
}