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.util.*;
import java.io.*;

public class Solution{
	long rem = 998244353L;
	
	public Solution(){
		Scanner sc = new Scanner(System.in);
		int tests = 1;
        
        for(int t=1; t<=tests; t++){
			int n = sc.nextInt();
			int m = sc.nextInt();
			int[] trees = new int[n];
			HashMap<Integer, Boolean> map = new HashMap<Integer, Boolean>();
			for(int i=0; i<n; i++){
				trees[i] = sc.nextInt();
				map.put(trees[i], true);
			}
			
			LinkedList<Pair> list = new LinkedList<Pair>();
			
			for(int i=0; i<n; i++){
				int left = trees[i]-1;
				int right = trees[i]+1;
				if(!map.containsKey(left)){
					map.put(left, true);
					list.add(new Pair(left, 1));
				}
				if(!map.containsKey(right)){
					map.put(right, true);
					list.add(new Pair(right, 1));
				}
			}
			
			long res = 0;
			int[] people = new int[m];
			int index = 0;
			while(index < m){
				Pair val = list.removeFirst();
				people[index++] = val.x;
				res += val.y;
				int left = val.x-1;
				int right = val.x+1;
				if(!map.containsKey(left)){
					map.put(left, true);
					list.add(new Pair(left, val.y+1));
				}
				if(!map.containsKey(right)){
					map.put(right, true);
					list.add(new Pair(right, val.y+1));
				}
			}
			
			System.out.println(res);
			for(int i=0; i<(m-1); i++){
				System.out.print(people[i]+" ");
			}
			System.out.println(people[m-1]);
        }
	}
	
	public static void main(String[] args){
		new Solution();
	}
}
class Pair{
	public int x;
	public int y;
	
	public Pair(int x, int y){
		this.x = x;
		this.y = y;
	}
}