Problem: 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.

Solution (in JAVA):

Solution:
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1015A extends PrintWriter {
	CF1015A() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1015A o = new CF1015A(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] dd = new int[m + 2];
		while (n-- > 0) {
			int l = sc.nextInt();
			int r = sc.nextInt();
			dd[l]++;
			dd[r + 1]--;
		}
		int k = 0;
		for (int d = 0, x = 1; x <= m; x++) {
			d += dd[x];
			if (d == 0)
				k++;
		}
		println(k);
		for (int d = 0, x = 1; x <= m; x++) {
			d += dd[x];
			if (d == 0)
				print(x + " ");
		}
		println();
	}
}