You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?

Input

The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.

Output

The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.

Examples

Input

2
0 2
2 5


Output

1
2 

Input

5
0 3
4 2
4 8
8 10
7 7


Output

3
7 10 3
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;

public class D {


	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt();
		
		int[] a = new int[n];
		int[] b = new int[n];

		Integer[] idx = new Integer[n];
		
		for (int i = 0; i < n; i++) {
			idx[i] = i;
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
			if(a[i] > b[i]) {
				int t = a[i];
				a[i] = b[i];
				b[i] = t;
			}
		}
		
		Arrays.sort(idx, new Comparator<Integer>() {

			@Override
			public int compare(Integer o1, Integer o2) {
				return b[o1] - b[o2];
			}
		});
		
		int l = -1000000;
		ArrayList<Integer> ans = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			int cur = idx[i];
			if(a[cur] > l) {
				ans.add(b[cur]);
				l = b[cur];
			}
		}
		
		out.println(ans.size());
		for (int i = 0; i < ans.size(); i++) 
			out.print(ans.get(i) + " ");
		out.println();
		
		out.flush();
		out.close();
	}


	static class Scanner 



	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}
		public Scanner(String file) throws FileNotFoundException{	br = new BufferedReader(new FileReader(file));}

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

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

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

		public String nextLine() throws IOException {return br.readLine();}


		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}

		public boolean ready() throws IOException {return br.ready();}



	}


}