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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Problem1015A {

	public static void main(String[] args) throws Exception {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);

		int n = in.nextInt();
		int m = in.nextInt();
		int[][] a = in.nextIntMatrix(n, 2);

		int[] b = new int[m + 1];

		int counter = 0;
		LOOP: for (int i = 1; i <= m; i++) {
			for (int j = 0; j < n; j++) {
				if (a[j][0] <= i && a[j][1] >= i) {
					continue LOOP;
				}
			}
			b[i] = 1;
			counter++;
		}

		out.println(counter);
		if (counter > 0) {
			for (int i = 1; i <= m; i++) {
				if (b[i] == 1) {
					out.print(i + " ");
				}
			}
		}

		out.close();
		return;
	}

	static long conv(long num, long base) {
		long res = 0;
		long p = 0;
		while (num != 0) {
			res += p * (num % base);
			num /= base;
			p = p * 10;
		}

		return res;
	}

	static class InputReader {
		BufferedReader reader;

		StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		private String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine(), " \t\n\r\f,");
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public int[] nextIntArray(int n) {
			return nextIntArray(n, 0);
		}

		public int[] nextIntArray(int n, int offset) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = nextInt() + offset;
			}
			return a;
		}

		public int[][] nextIntMatrix(int n, int m) {
			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 String nextLine() {
			return next();
		}

		public char[] nextLineCharArray() {
			return next().toCharArray();
		}
	}

}