Problem: This is the easy version of the problem. The difference between versions is the constraints on n and a_i. You can make hacks only if all versions of the problem are solved.

First, Aoi came up with the following idea for the competitive programming problem:

Yuzu is a girl who collecting candies. Originally, she has x candies. There are also n enemies numbered with integers from 1 to n. Enemy i has a_i candies.

Yuzu is going to determine a permutation P. A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, \{2,3,1,5,4\} is a permutation, but \{1,2,2\} is not a permutation (2 appears twice in the array) and \{1,3,4\} is also not a permutation (because n=3 but there is the number 4 in the array).

After that, she will do n duels with the enemies with the following rules:

  * If Yuzu has equal or more number of candies than enemy P_i, she wins the duel and gets 1 candy. Otherwise, she loses the duel and gets nothing. 
  * The candy which Yuzu gets will be used in the next duels. 



Yuzu wants to win all duels. How many valid permutations P exist?

This problem was easy and wasn't interesting for Akari, who is a friend of Aoi. And Akari made the following problem from the above idea:

Let's define f(x) as the number of valid permutations for the integer x.

You are given n, a and a prime number p ≤ n. Let's call a positive integer x good, if the value f(x) is not divisible by p. Find all good integers x.

Your task is to solve this problem made by Akari.

Input

The first line contains two integers n, p (2 ≤ p ≤ n ≤ 2000). It is guaranteed, that the number p is prime (it has exactly two divisors 1 and p).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2000).

Output

In the first line, print the number of good integers x.

In the second line, output all good integers x in the ascending order.

It is guaranteed that the number of good integers x does not exceed 10^5.

Examples

Input


3 2
3 4 5


Output


1
3


Input


4 3
2 3 5 6


Output


2
3 4


Input


4 3
9 1 1 1


Output


0

Note

In the first test, p=2.

  * If x ≤ 2, there are no valid permutations for Yuzu. So f(x)=0 for all x ≤ 2. The number 0 is divisible by 2, so all integers x ≤ 2 are not good. 
  * If x = 3, \{1,2,3\} is the only valid permutation for Yuzu. So f(3)=1, so the number 3 is good. 
  * If x = 4, \{1,2,3\} , \{1,3,2\} , \{2,1,3\} , \{2,3,1\} are all valid permutations for Yuzu. So f(4)=4, so the number 4 is not good. 
  * If x ≥ 5, all 6 permutations are valid for Yuzu. So f(x)=6 for all x ≥ 5, so all integers x ≥ 5 are not good. 



So, the only good number is 3.

In the third test, for all positive integers x the value f(x) is divisible by p = 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.*;
import java.util.*;
public class AsterismEZ { 
 
	public static void main(String[] args) { 
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		int N = I.nextInt();
		int p = I.nextInt();
		int[] cnts = new int[2001]; //Number of numbers <= i in array
		int[] a = new int[N];
		for (int i = 0; i < N; i++) a[i]=I.nextInt();
		for (int i = 0; i < N; i++) cnts[a[i]]++; 
		for (int i = 2; i <= 2000; i++) cnts[i]+=cnts[i-1];
		
		Queue<Integer> ans = new LinkedList<Integer>();
		for (int x = 1; x <= 2000; x++) {
			boolean good = true;
			int curCandy = x; //Current number of candies
			int used = 0; //Number of candies slotted so far
			for (int test = 0; test < N; test++) {
				if ((cnts[curCandy]-used)%p==0) {
					good=false;
					break;
				} 
				used++; //Ultimately, and if possible, one candy will have been used
				curCandy++;
				curCandy = Math.min(2000, curCandy); //a[i] is guaranteed to be <= 2000
			}
			if (good) ans.add(x);
		}
		O.pln(ans.size());
		while (!ans.isEmpty()) O.p(ans.poll()+" ");
	}
	public static long ceil(long num, long den) {long ans = num/den; if (num%den!=0) 
	ans++; return ans;}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return Math.max(a,b);
		return GCD(Math.min(a, b),Math.max(a, b)%Math.min(a, b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());};
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void pln(StringBuilder x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
	}
}