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 Main {
	static Scanner scanner = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	static int n,m, arr[];
	static long sum = 0;
	static ArrayList<Integer> ans = new ArrayList<>();
	
	static void bfs() {
		HashMap<Integer,Integer> dist = new HashMap<>();
		Queue<Integer> q = new LinkedList<>();
		
		for(int i=0; i<n; i++) {
			dist.put(arr[i],0);
			q.add(arr[i]);
		}
		
		while(!q.isEmpty() && m > 0) {
			int u = q.poll();
			int d = dist.get(u);
			
			if(!dist.containsKey(u - 1)) {
				q.add(u-1);
				dist.put(u-1,d + 1);
				ans.add(u -1);
				sum += d + 1;
				m--;
			}
			
			if(m > 0 && !dist.containsKey(u + 1)) {
				q.add(u + 1);
				ans.add(u + 1);
				dist.put(u + 1 , d + 1);
				sum += d + 1;
				m--;
			}
		}
		
	}
	
	
	public static void main(String[] args) throws Exception {
		n = scanner.nextInt();
		m = scanner.nextInt();
		arr = new int[n];
		
		for(int i=0; i<n; i++) {
			arr[i]= scanner.nextInt();
		}
		
		bfs();
		out.println(sum);
		for(int x : ans) {
			out.print(x + " ");
		}
		out.close();
	}

}

class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream system) {
		br = new BufferedReader(new InputStreamReader(system));
	}

	public Scanner(String file) throws Exception {
		br = new BufferedReader(new FileReader(file));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public String nextLine() throws IOException {
		return br.readLine();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public Long nextLong() throws IOException {
		return Long.parseLong(next());
	}

}