Toad Ivan has m pairs of integers, each integer is between 1 and n, inclusive. The pairs are (a_1, b_1), (a_2, b_2), …, (a_m, b_m). 

He asks you to check if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 300 000, 1 ≤ m ≤ 300 000) — the upper bound on the values of integers in the pairs, and the number of given pairs.

The next m lines contain two integers each, the i-th of them contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the integers in the i-th pair.

Output

Output "YES" if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y. Otherwise, print "NO". You can print each letter in any case (upper or lower).

Examples

Input


4 6
1 2
1 3
1 4
2 3
2 4
3 4


Output


NO


Input


5 4
1 2
2 3
3 4
4 5


Output


YES


Input


300000 5
1 2
1 2
1 2
1 2
1 2


Output


YES

Note

In the first example, you can't choose any x, y because for each such pair you can find a given pair where both numbers are different from chosen integers.

In the second example, you can choose x=2 and y=4.

In the third example, you can choose x=1 and y=2.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main
{
	public static void main(String[] args)
	{
		FastReader fr =new FastReader();

		PrintWriter op =new PrintWriter(System.out);

		int n =fr.nextInt() ,m =fr.nextInt() ,pair[][] =new int[m][2] ,frq[] =new int[n+1] ,i ,j ,ctr =1 ;

		j =pair[0][0] =fr.nextInt() ;	pair[0][1] =fr.nextInt() ;	boolean f =false ;

		for (i =1 ; i<m ; i++) {
			pair[i][0] =fr.nextInt() ;	pair[i][1] =fr.nextInt() ;

			if (pair[i][0] != j && pair[i][1] != j)	{	frq[pair[i][0]]++ ;	frq[pair[i][1]]++ ;		}
			else ++ctr ;
		}

		for (i =1 ; i<=n ; i++) {
			if (frq[i] == m-ctr)	{f =true ;break;}
		}
		if (!f) {
			j =pair[0][1] ;		ctr =1 ;	frq =new int[n+1] ;

			for (i =1 ; i<m ; i++) {
				if (pair[i][0] != j && pair[i][1] != j)	{	frq[pair[i][0]]++ ;	frq[pair[i][1]]++ ;		}
				else ++ctr ;
			}
			for (i =1 ; i<=n ; i++) {
				if (frq[i] == m-ctr)	{f =true ;break;}
			}
		}

		if (f)	op.println("YES") ;
		else op.println("NO") ;
		op.flush();	op.close();
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br =new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st==null || (!st.hasMoreElements())) 
			{
				try
				{
					st =new StringTokenizer(br.readLine());
				}
				catch(IOException e)
				{
					e.printStackTrace();
				}
				
			}
			return st.nextToken();
		}

		String nextLine() {
			String str ="";

			try
			{
				str =br.readLine();
			}
			catch(IOException e)
			{
				e.printStackTrace();
			}

			return str;
		}

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

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