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.util.*;
public class Pairs562B {
	static class Pair {
		int left,right;
		public Pair(int left,int right) {
			this.left= left;
			this.right = right;
		}
	
	}
	public static void main(String[] args) {
		Scanner s =new Scanner(System.in);
		int n = s.nextInt();
		int m = s.nextInt();
		Pair pair[] = new Pair[m];
		int count[] = new int[n+1];
		for(int i=0;i<m;i++) {
			int l = s.nextInt();
			int r = s.nextInt();
			pair[i] = new Pair(l,r);
		}
		Pair pr = pair[0];
		int x = pr.left;
		
		for(Pair p:pair) {
			if(p.left==x||p.right==x) {
			count[x]++;
			}else {
				count[p.left]++;
				count[p.right]++;
			}
		}
		x = count[x];
		int y = findMax(count,pr.left);
		
		if(x+y==m) {
			System.out.println("YES");
			return;
		}
		Arrays.fill(count, 0);
		 x = pr.right;
		 for(Pair p:pair) {
				if(p.left==x||p.right==x) {
				count[x]++;
				}else {
					count[p.left]++;
					count[p.right]++;
				}
			}
			x = count[x];
			y = findMax(count,pr.right);
			
			if(x+y==m) {
				System.out.println("YES");
				return;
			}
			System.out.println("NO");

		
		
	}
	private static int findMax(int[] right,int notthis) {
		int max = Integer.MIN_VALUE;
		int index = 0;
		for(int i =0;i<right.length;i++) {
			if(right[i]>max && i!=notthis) {
				max = right[i];
				index = i;
			}
		}
		return max;
	}
	


}
