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 Div3611;

import javafx.util.Pair;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class ProblemD {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String line[]=br.readLine().split(" ");
        int n=Integer.parseInt(line[0]);
        int m=Integer.parseInt(line[1]);
        line=br.readLine().split(" ");
        long a[]=new long[n];
        HashMap<Long,Integer> marked=new HashMap<>();
        Queue<Pair<Long,Long>> q=new LinkedList<>();
        for(int i=0;i<n;i++){
            a[i]=Long.parseLong(line[i]);
            marked.put(a[i],1);
        }
        for(int i=0;i<n;i++){
            if(!marked.containsKey(a[i]+1)){
                q.add(new Pair<>(a[i]+1,1l));
                marked.put(a[i]+1,1);
            }
            if(!marked.containsKey(a[i]-1)){
                q.add(new Pair<>(a[i]-1,-1l));
                marked.put(a[i]-1,1);
            }
        }
        ArrayList<Long> ans=new ArrayList<>();
        long total=0;
        int count=0;
        while(!q.isEmpty()){
            Pair<Long,Long> curr=q.remove();
            long y=curr.getValue();
            long x=curr.getKey();
            if(y>0){
                total+=y;
                ans.add(x);
                count++;
                if(!marked.containsKey(x+1)){
                    q.add(new Pair<>(x+1,y+1));
                    marked.put(x+1,1);
                }
            }
            else{
                total-=y;
                ans.add(x);
                count++;
                if(!marked.containsKey(x-1)){
                    q.add(new Pair<>(x-1,y-1));
                    marked.put(x-1,1);
                }
            }
            if(count==m){
                break;
            }
        }
        System.out.println(total);
        StringBuilder str=new StringBuilder();
        for(int i=0;i<ans.size();i++){
            str.append(ans.get(i)+" ");
        }
        System.out.println(str.toString());
    }
}
