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.Scanner;

public class PairsSolution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        String upperBoundAndCount = in.nextLine();
        int m = Integer.parseInt(upperBoundAndCount.split(" ")[1]);

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

        for(int i = 0; i < m; i++) {
            String[] parts = in.nextLine().split(" ");
            pairs[i][0] = Integer.parseInt(parts[0]);
            pairs[i][1] = Integer.parseInt(parts[1]);
        }
    
        String result = pairExists(pairs) ? "YES" : "NO";
        System.out.println(result);
    }

    static boolean pairExists(int[][] pairs) {
        int[] firsts = new int[]{pairs[0][0], pairs[0][1]};
        for (int first : firsts) {
            if (coversAll(pairs, first, -1)) {
                return true;
            }

            for (int i = 1; i < pairs.length; i++) {
                int[] pair = pairs[i];
                if (pair[0] != first && pair[1] != first) {
                    if (coversAll(pairs, first, pair[0]) || coversAll(pairs, first, pair[1])) {
                        return true;
                    }
                    break;
                }
            }
        }

        return false;
    }

    static boolean coversAll(int[][] pairs, int first, int second) {
        for (int[] pair : pairs) {
            if (pair[0] != first && pair[0] != second && pair[1] != first && pair[1] != second) {
                return false;
            }
        }
        return true;
    }
}
