There are n Christmas trees on an infinite number line. The i-th tree grows at the position x_i. All x_i are guaranteed to be distinct.

Each integer point can be either occupied by the Christmas tree, by the human or not occupied at all. Non-integer points cannot be occupied by anything.

There are m people who want to celebrate Christmas. Let y_1, y_2, ..., y_m be the positions of people (note that all values x_1, x_2, ..., x_n, y_1, y_2, ..., y_m should be distinct and all y_j should be integer). You want to find such an arrangement of people that the value ∑_{j=1}^{m}min_{i=1}^{n}|x_i - y_j| is the minimum possible (in other words, the sum of distances to the nearest Christmas tree for all people should be minimized).

In other words, let d_j be the distance from the j-th human to the nearest Christmas tree (d_j = min_{i=1}^{n} |y_j - x_i|). Then you need to choose such positions y_1, y_2, ..., y_m that ∑_{j=1}^{m} d_j is the minimum possible.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of Christmas trees and the number of people.

The second line of the input contains n integers x_1, x_2, ..., x_n (-10^9 ≤ x_i ≤ 10^9), where x_i is the position of the i-th Christmas tree. It is guaranteed that all x_i are distinct.

Output

In the first line print one integer res — the minimum possible value of ∑_{j=1}^{m}min_{i=1}^{n}|x_i - y_j| (in other words, the sum of distances to the nearest Christmas tree for all people).

In the second line print m integers y_1, y_2, ..., y_m (-2 ⋅ 10^9 ≤ y_j ≤ 2 ⋅ 10^9), where y_j is the position of the j-th human. All y_j should be distinct and all values x_1, x_2, ..., x_n, y_1, y_2, ..., y_m should be distinct.

If there are multiple answers, print any of them.

Examples

Input


2 6
1 5


Output


8
-1 2 6 4 0 3 


Input


3 5
0 3 1


Output


7
5 -2 4 -1 2 
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Map;
import java.util.Collection;
import java.util.Scanner;
import java.util.HashMap;
import java.util.Queue;
import java.util.LinkedList;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Kraken7
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    Scanner in = new Scanner(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    TaskD solver = new TaskD();
    solver.solve(1, in, out);
    out.close();
  }

  static class TaskD {

    public void solve(int testNumber, Scanner in, PrintWriter out) {
      int n = in.nextInt(), m = in.nextInt();
      Queue<Long> q = new LinkedList<>();
      Map<Long, Long> dis = new HashMap<>();
      for (int i = 0; i < n; i++) {
        long x = in.nextLong();
        q.add(x);
        dis.put(x, 0L);
      }
      ArrayList<Long> res = new ArrayList<>();
      long ans = 0;
      while (!q.isEmpty() && m > 0) {
        long curr = q.poll();
        if (dis.get(curr) != 0) {
          res.add(curr);
          ans += dis.get(curr);
          m--;
        }
        if (!dis.containsKey(curr - 1)) {
          dis.put(curr - 1, dis.get(curr) + 1);
          q.add(curr - 1);
        }
        if (!dis.containsKey(curr + 1)) {
          dis.put(curr + 1, dis.get(curr) + 1);
          q.add(curr + 1);
        }
      }
      out.println(ans);
      for (long i : res)
        out.print(i + " ");
    }

  }
}

