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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Set;
import java.util.StringTokenizer;

public class TaskD {

	public static void main(String[] arg) {
		final FastScanner in = new FastScanner(System.in);
		final PrintWriter out = new PrintWriter(System.out);
		final int n = in.nextInt();
		final int k = in.nextInt();
		final int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
		}
		out.println("" + solution(n, a, k));
		out.close();
		in.close();
	}

	private static long solution(int n, int[] a, int k) {
		// brute forse
		Set<Integer> possibleNumbers = new HashSet<>();
		possibleNumbers.add(0);

		Map<Integer, PriorityQueue<Integer>> map = new HashMap<>();
		for (int i = 0; i < n; i++) {
			int num = a[i];
			possibleNumbers.add(num);
			int count = 0;
			{
				PriorityQueue<Integer> pq = map.get(num);
				if (pq == null) {
					pq = new PriorityQueue<Integer>();
					map.put(num, pq);
				}
				pq.add(count);
			}
			while (num > 0) {
				num >>= 1;
				possibleNumbers.add(num);
				count++;

				PriorityQueue<Integer> pq = map.get(num);
				if (pq == null) {
					pq = new PriorityQueue<Integer>();
					map.put(num, pq);
				}
				pq.add(count);
			}
		}
		Long minCount = Long.MAX_VALUE;

		for (Integer num : possibleNumbers) {
			// k equals numbers
			PriorityQueue<Integer> pq = map.get(num);
			if (pq != null && pq.size() >= k) {
				long midRes = 0;
				for (int i = 0; i < k; i++) {
					midRes += pq.poll();
				}
				minCount = Math.min(minCount, midRes);
			}
		}
		return minCount;
	}

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner(InputStream stream) {
			try {
				br = new BufferedReader(new InputStreamReader(stream));
			} catch (Exception e) {
				e.printStackTrace();
			}
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}

	}
}