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.*;
import java.util.*;
public class Solution{
	public static void main(String args[]){
		Scanner sc=new Scanner(System.in);
		int n =sc.nextInt();
		int m=sc.nextInt();
		int[] a=new int[m];
		int b[]=new int[m];
		for(int i=0;i<m;i++){
			a[i]=sc.nextInt();
			b[i]=sc.nextInt();
		}
		
		System.out.println(foo(n,m,a,b));
		
	}
	public static String foo(int n,int m,int[] a,int[] b){
		int[] freq=new int[300001];
		int[] x=new int[2];
		x[0]=a[0];
		x[1]=b[0];
		for(int xi:x){
			int cnt=0;
			for(int i=1;i<m;i++){
				if(a[i]!=xi && b[i]!=xi){
					freq[a[i]]++;
					freq[b[i]]++;
					cnt++;
				}
			}
			//System.out.println(cnt+" "+freq[2]+" "+freq[3]+" //"+freq[4]);
			for(int i=1;i<m;i++){
				if(freq[a[i]]==cnt || freq[b[i]]==cnt)
					return "YES";
			}
			Arrays.fill(freq,0);
		}
		if(m==1)
			return "YES";
		return "NO";
		
		
	}
}
	