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 Pairs {
	
	static int[][] p ;

	public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt(), m = in.nextInt();
        int freq[] = new int[n+1];
        p = new int[m][2];
        for(int i=0;i<m;i++) {
        	int x = in.nextInt();
        	freq[x]++;
        	int y = in.nextInt();
        	freq[y]++;
        	p[i][0] = x;
        	p[i][1] = y;
        }
        boolean verdict = false;
        ArrayList<Integer> ansx = new ArrayList<>();
        for(int i=1;i<=n;i++) {
    		if(freq[i]>=(m+1)/2) ansx.add(i);
    	}
    	for(int i : ansx) {
    		int y1,y2;y1=y2=-1;
    		for(int j=0;j<m;j++) {
    			if(p[j][0]!=i&&p[j][1]!=i) {
    				if(y1==-1&&y2==-1) {
    					y1 = p[j][0];y2 = p[j][1];
    				}
    				else {
    					if(p[j][0]!=y1&&p[j][1]!=y1) y1 = 0;
    					if(p[j][0]!=y2&&p[j][1]!=y2) y2 = 0;
    				}
    				if(y1==0&&y2==0)
    					break;
    			}
    		}
    		if(y1!=0||y2!=0) {
    			verdict = true; break;
    		}
    	}
        if(verdict)
        	out.println("YES");
        else
        	out.println("NO");
        out.flush();
	}
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		String next() {
			while(!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				}catch(IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int a[] = new int[n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
		
		Integer[] readArray(long n) {
		    Integer a[] = new Integer[(int)n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
		
		long[] readArray(int n,long x) {
			long a[] = new long[n];
			for(int i=0;i<n;i++) a[i] = nextLong();
			return a;
		}
		
	}
}
