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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.HashSet;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class D {
	static PrintWriter out = new PrintWriter(System.out);

	
	public static void main(String[] args) {
		FS in = new FS();

		int N = in.nextInt();
		int M = in.nextInt();
		TreeSet<Long> trees = new TreeSet<Long>();
		for(int i = 0; i < N; i++) {
			trees.add(in.nextLong());
		}
		
		TreeSet<Long> ps = new TreeSet<Long>();
		
		HashSet<Long> vis = new HashSet<Long>();
		ArrayDeque<Long> q = new ArrayDeque<Long>();
		for(long ll : trees) {
			if(!vis.contains(ll+1) && !trees.contains(ll+1)) q.add(ll+1);
			if(!vis.contains(ll-1) && !trees.contains(ll-1)) q.add(ll-1);
			vis.add(ll+1);
			vis.add(ll-1);
		}
		
		while(ps.size() < M) {
			long pos = q.poll();
			if(ps.contains(pos)) continue;
			ps.add(pos);
			if(!vis.contains(pos-1) && !trees.contains(pos-1)) q.add(pos-1);
			if(!vis.contains(pos+1)&& !trees.contains(pos+1)) q.add(pos+1);
			vis.add(pos-1);
			vis.add(pos+1);
		}
		
		long res = 0;
		for(long ll : ps) {
			long val = Long.MAX_VALUE;
			if(trees.lower(ll) != null) val = Math.min(val,ll-trees.lower(ll));
			if(trees.higher(ll) != null) val = Math.min(val, trees.higher(ll)-ll);
			res += val;
		}
		
		out.println(res);
		for(long ll : ps) out.print(ll+" ");
		out.println();
		
	
		out.close();
	}
	
	
	static class FS{
		BufferedReader br;
		StringTokenizer st;
		public FS() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch(Exception e) { throw null;}
			}
			return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next());}
		double nextDouble() { return Double.parseDouble(next());}
		long nextLong() { return Long.parseLong(next());}
		int[] NIA(int n) {
			int r[] = new int[n];
			for(int i = 0; i < n; i++) r[i] = nextInt();
			return r;
		}
		long[] NLA(int n) {
			long r[] = new long[n];
			for(int i = 0; i < n; i++) r[i] = nextLong();
			return r;
		}
		char[][] grid(int r, int c){
			char res[][] = new char[r][c];
			for(int i = 0; i < r; i++) {
				char l[] = next().toCharArray();
				for(int j = 0; j < c; j++) {
					res[i][j] = l[j];
				}
			}
			return res;
		}
	}
	
}
