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.IOException;
import java.io.InputStream;
import java.util.InputMismatchException;

public class P1015A
{
	public static void main(String[] args)
	{
		FastScanner in = new FastScanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		boolean[] covered = new boolean[m + 1];
		while (n-->0) {
			int l = in.nextInt();
			int r = in.nextInt();
			for (int i = l; i <= r; i++)
				covered[i] = true;
		}
		int count = 0;
		for (int i = 1; i <= m; i++)
			if (!covered[i])
				count++;
		System.out.println(count);
		for (int i = 1; i <= m; i++)
			if (!covered[i])
				System.out.print(i + " ");
	}
	
	/**
	 * Source: Matt Fontaine
	 */
	public static class FastScanner
	{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public FastScanner(InputStream stream)
		{
			this.stream = stream;
		}

		int read()
		{
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars)
			{
				curChar = 0;
				try
				{
					numChars = stream.read(buf);
				} catch (IOException e)
				{
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c)
		{
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c)
		{
			return c == '\n' || c == '\r' || c == -1;
		}

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

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

		public double nextDouble()
		{
			return Double.parseDouble(next());
		}

		public String next()
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine()
		{
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
	}
}