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 
/*
 * Created by cravuri on 12/28/19
 */

import java.util.*;

public class ChristmasTrees {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long[] x = new long[n];
        Set<Long> treeLoc = new HashSet<>();
        for (int i = 0; i < n; i++) {
            x[i] = sc.nextLong();
            treeLoc.add(x[i]);
        }

        long res = 0;
        List<int[]> trees = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            trees.add(new int[] {i, 1});
            trees.add(new int[] {i, -1});
        }
        Set<Long> people = new HashSet<>();
        for (long dis = 1; ; dis++) {
            List<int[]> newTrees = new ArrayList<>();
            for (int i = 0; i < trees.size(); i++) {
                long pos;
                if (trees.get(i)[1] == 1) {
                    pos = x[trees.get(i)[0]] + dis;
                } else {
                    pos = x[trees.get(i)[0]] - dis;
                }

                if (!people.contains(pos) && !treeLoc.contains(pos)) {
                    newTrees.add(trees.get(i));
                    people.add(pos);
                    res += dis;
                    if (people.size() >= m) {
                        break;
                    }
                }
//                int left = x[trees.get(i)[0]] - dis;
//                int right = x[trees.get(i)[0]] + dis;
//                if (!people.contains(left) || !people.contains(right)) {
//                    newTrees.add(trees.get(i));
//                    if (!people.contains(left) && !treeLoc.contains(left)) {
//                        people.add(left);
//                        res += dis;
//                        if (people.size() >= m) {
//                            break;
//                        }
//                    }
//                    if (!people.contains(right) && !treeLoc.contains(right)) {
//                        people.add(right);
//                        res += dis;
//                        if (people.size() >= m) {
//                            break;
//                        }
//                    }
//                }
            }

            if (people.size() >= m) {
                break;
            }

            trees.clear();
            trees.addAll(newTrees);
        }

        System.out.println(res);
        for (Long p : people) {
            System.out.print(p + " ");
        }
        System.out.println();
    }

}
