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 
//package circuits;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.Scanner;

public class bruteforce {
  static class Node {
    int index;
    int cost;
    public Node(int index, int cost) {
      this.index = index;
      this.cost = cost;
    }
  }
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int m = sc.nextInt();
      int data[] = new int[n];
      LinkedList<Node> q = new LinkedList<>();
      Map<Integer, Integer> visited = new HashMap<>();
    for (int i = 0; i < n; i++) {
      data[i] = sc.nextInt();
      q.add(new Node(data[i], 0));
      visited.put(data[i], 1);
    }
    int count = 0;
    long ans_value = 0;
    StringBuilder ans = new StringBuilder();
    while (count < m && q.size() != 0) {
      Node curr = q.poll();
      if(!visited.containsKey(curr.index-1) && curr.index-1 >= -2000000000) {
        visited.put(curr.index - 1, 1);
        count++;
        q.add(new Node(curr.index-1, curr.cost+1));
        ans_value += curr.cost+1;
        ans.append(curr.index-1).append(" ");
      }
      if(!visited.containsKey(curr.index+1) && count < m && curr.index+1 <= 2000000000) {
        visited.put(curr.index + 1, 1);
        count++;
        q.add(new Node(curr.index+1, curr.cost+1));
        ans_value += curr.cost + 1;
        ans.append(curr.index+1).append(" ");
      }
    }
    System.out.println(ans_value);
    System.out.println(ans.toString());

  }

}
