You are given a set of n segments on the axis Ox, each segment has integer endpoints between 1 and m inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — coordinates of the left and of the right endpoints. 

Consider all integer points between 1 and m inclusive. Your task is to print all such points that don't belong to any segment. The point x belongs to the segment [l; r] if and only if l ≤ x ≤ r.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100) — the number of segments and the upper bound for coordinates.

The next n lines contain two integers each l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — the endpoints of the i-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that l_i=r_i, i.e. a segment can degenerate to a point.

Output

In the first line print one integer k — the number of points that don't belong to any segment.

In the second line print exactly k integers in any order — the points that don't belong to any segment. All points you print should be distinct.

If there are no such points at all, print a single integer 0 in the first line and either leave the second line empty or do not print it at all.

Examples

Input

3 5
2 2
1 2
5 5


Output

2
3 4 


Input

1 7
1 7


Output

0

Note

In the first example the point 1 belongs to the second segment, the point 2 belongs to the first and the second segments and the point 5 belongs to the third segment. The points 3 and 4 do not belong to any segment.

In the second example all the points from 1 to 7 belong to the first segment.
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class PointsInSegments {
	public static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	public static StringTokenizer st;
	
	public static void main(String[] args) throws IOException {
		int n = nextInt();
		int m = nextInt();
		boolean[] a = new boolean[m+1];
		for (int i = 0; i < n; i++) {
			int l = nextInt();
			int r = nextInt();
			for (int j = l; j <= r; j++)
				a[j] = true;
		}
		
		int k = 0;
		String s = "";
		for (int i = 1; i <= m; i++)
			if (!a[i]) {
				k++;
				s += " " + i;
			}
		
		System.out.println(k);
		System.out.println(s.length() == 0 ? s : s.substring(1));
	}
	
	public static String nextLine() throws IOException {
		return in.readLine();
	}

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

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

	public static long nextLong() throws IOException {
		return Long.parseLong(nextString());
	}

	public static int[] intArray(int n) throws IOException {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public static int[][] intArray(int n, int m) throws IOException {
		int[][] a = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				a[i][j] = nextInt();
		return a;
	}

	public static long[] longArray(int n) throws IOException {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}