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.*;
import java.util.*;

public class trees {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		
		StringTokenizer st = new StringTokenizer(br.readLine());
		int tree = Integer.parseInt(st.nextToken());
		int people = Integer.parseInt(st.nextToken());
		
		TreeSet<Integer> ts = new TreeSet<Integer>();
		int[] x = new int[tree];
		int[] y = new int[people];
		
		long dist = 0;
		st = new StringTokenizer(br.readLine());
		for (int i = 0; i < tree; i++) ts.add(x[i] = Integer.parseInt(st.nextToken()));
		
		LinkedList<State> l = new LinkedList<State>();
		TreeSet<Integer> used = new TreeSet<Integer>();
		int tot = 0;
		
		for (int i = 0; i < tree; i++){
			if (!ts.contains(x[i] - 1) && !used.contains(x[i] - 1)) {
				l.add(new State(x[i] - 1, 1, 1));
				used.add(x[i] - 1);
			}
			if (!ts.contains(x[i] + 1) && !used.contains(x[i] + 1)) {
				l.add(new State(x[i]  + 1, 1, -1));
				used.add(x[i] + 1);
			}
		}
		
		while (tot != people){
			State s = l.poll();
			dist = dist + (long) s.dist;
			
			y[tot] = s.loc;
			tot++;
			
			if (s.left == 1 && !ts.contains(s.loc - 1) && !used.contains(s.loc - 1)){
				l.add(new State(s.loc - 1, s.dist + 1, 1));
				used.add(s.loc - 1);
			}
			if (s.left == -1 && !ts.contains(s.loc + 1) && !used.contains(s.loc + 1)){
				l.add(new State(s.loc + 1, s.dist + 1, -1));
				used.add(s.loc + 1);
			}
		}
		
		pw.println(dist);
		for (int i = 0; i < people; i++) pw.print(y[i] + " ");
		
		pw.println();
		pw.close();
		br.close();
	}
	
	// left == 1, right == -1
	static class State {
		public int loc, dist, left;
		public State(int a, int b, int c){
			loc = a;
			dist = b;
			left = c;
		}
	}
}
