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{
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String st[] = br.readLine().split(" ");

		int n = Integer.parseInt(st[0]);
		int m = Integer.parseInt(st[1]);

		int input[][] = new int[m][2];

		int i;

		for(i = 0; i < m; i++){
			st = br.readLine().split(" ");

			input[i][0] = Integer.parseInt(st[0]);
			input[i][1] = Integer.parseInt(st[1]);
		}

		int c1 = 1;
		int c2 = 1;

		int count[][] = new int[2][n+1];

		int a = input[0][0];
		int b = input[0][1];

		for(i = 1; i < m; i++){
			if((a == input[i][0]) || (a == input[i][1])){
				c1++;
			}
			else{
				count[0][input[i][0]] += 1;
				count[0][input[i][1]] += 1; 
			}

			if((b == input[i][0]) || (b == input[i][1])){
				c2++;
			}
			else{
				count[1][input[i][0]] += 1;
				count[1][input[i][1]] += 1; 
			}
		}

		boolean flag = false;

		for(i = 1; i <= n; i++){
			if((c1 + count[0][i]) == m){
				flag = true;
				break;
			}

			if((c2 + count[1][i]) == m){
				flag = true;
				break;
			}
		}

		if(flag){
			System.out.println("YES");
		}
		else{
			System.out.println("NO");
		}
	}
}