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 b {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();

        int[] freq = new int[n + 1];
        int[] pairs = new int[m * 2];
        int offset = 0;
        for (int i = 0; i < m; ++i) {
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            pairs[offset] = a;
            pairs[offset + 1] = b;
            offset += 2;

            freq[a]++;
            freq[b]++;
        }
        scanner.close();

//        int maxFreq = freq[1];
//        for (int i = 2; i <= n; ++i) {
//            if (freq[i] > maxFreq) {
//                maxFreq = freq[i];
//            }
//        }
//
//        if (maxFreq < m / 2) {
//            System.out.println("NO");
//            return;
//        }

        for (int maxVal = 1; maxVal <= n; ++maxVal) {
            if (freq[maxVal] < m/2/* && freq[maxVal] < maxFreq*/) {
                continue;
            }

            offset = 0;
            Integer x = 0;
            Integer y = 0;
            for (int i = 0; i < m; ++i) {
                int a = pairs[offset];
                int b = pairs[offset + 1];
                offset += 2;

                if (a == maxVal || b == maxVal) {
                    continue;
                }

                if (x != null && x == 0) {
                    x = a;
                    y = b;
                } else {
                    if (x != null && x != a && x != b) {
                        x = null;
                    }

                    if (y != null && y != a && y != b) {
                        y = null;
                    }

                    if (x == null && y == null) {
                        break;
                    }
                }
            }


            if (x != null || y != null) {
                System.out.println("YES");
                return;
            }
        }

        System.out.println("NO");
    }
    
}
