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.*;
import java.util.concurrent.TimeUnit;

public class tree {
	
	public static void main (String [] args) throws Exception {
	    BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	    StringTokenizer st = new StringTokenizer(f.readLine());
	    int N = Integer.parseInt(st.nextToken());
	    int M = Integer.parseInt(st.nextToken());
	    
	    Set<Long> trees = new HashSet<>();
	    Set<Long> people = new HashSet<>();
	    Queue<long[]> queue = new LinkedList<long[]>();
	    st = new StringTokenizer(f.readLine());
	    for(int i = 0; i < N; i++) {
	    	long t = Long.parseLong(st.nextToken());
	    	trees.add(t);
	    	queue.add(new long[] {t+1, 1});
	    	queue.add(new long[] {t-1, -1});
	    }
	    
	    long sum = 0;
	    while(true) {
	    	if(people.size() >= M)
	    		break;
	    	long[] stor = queue.poll();
	    	if(!people.contains(stor[0]) && !trees.contains(stor[0])) {
	    		//System.out.println(stor[0] + " " + trees);
		    	people.add(stor[0]);
		    	sum += Math.abs(stor[1]);
		    	//System.out.println(stor[0] + " " + stor[1]);
		    	if(stor[1] > 0) {
		    		queue.add(new long[] {stor[0] + 1, stor[1] + 1});
		    	} else {
		    		queue.add(new long[] {stor[0] - 1, stor[1] - 1});
		    	}
	    	}
	    }
	    
	    out.println(sum);
	    Iterator<Long> it = people.iterator();
	    long x = it.next();
	    out.print(x);
	    while(it.hasNext()) {
	    		out.print(" " + it.next());
	   	}
	    out.println();
	    
	    out.close();
	    f.close();
	}
	
}