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.*;

public class _1283D {

    Scanner scanner = new Scanner(System.in);
    int n, m;
    int[] christmasTree;
    Set<Integer> occupied = new HashSet<>();
    LinkedList<BfsCoordinate> nextPossibleLocation = new LinkedList<>();
    long totalDistance;
    List<Integer> humanLocation = new ArrayList<>();

    private class BfsCoordinate {
        int xCoordinate;
        boolean forward;
        int distanceFromTree;

        public BfsCoordinate(int xCoordinate, boolean forward, int distanceFromTree) {
            this.xCoordinate = xCoordinate;
            this.forward = forward;
            this.distanceFromTree = distanceFromTree;
        }
    }

    public static void main(String[] args) {
        _1283D solution = new _1283D();
        solution.solve();
    }

    public void solve(){
        n = scanner.nextInt();
        m = scanner.nextInt();
        christmasTree = new int[n+1];

        for(int i=1; i<=n; i++){
            christmasTree[i] = scanner.nextInt();
            nextPossibleLocation.add(new BfsCoordinate(christmasTree[i], true, 0));
            nextPossibleLocation.add(new BfsCoordinate(christmasTree[i], false, 0));
            occupied.add(christmasTree[i]);
        }


        BfsCoordinate posLoc;
        int nextLoc;
        while(!nextPossibleLocation.isEmpty() && humanLocation.size() < m){
            posLoc = nextPossibleLocation.poll();
            if(posLoc.forward){
                nextLoc = posLoc.xCoordinate + 1;
            }
            else{
                nextLoc = posLoc.xCoordinate - 1;
            }

            if(!occupied.contains(nextLoc)){
                totalDistance += posLoc.distanceFromTree+1;
                humanLocation.add(nextLoc);
                occupied.add(nextLoc);
                nextPossibleLocation.addLast(new BfsCoordinate(nextLoc, posLoc.forward, posLoc.distanceFromTree+1));
            }
        }

        System.out.println(totalDistance);
        for(Integer loc: humanLocation){
            System.out.print(loc + " ");
        }
    }

}