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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author aryssoncf
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            try {
                int n = in.nextInt(), m = in.nextInt();
                int[] A = new int[m], B = new int[m], freq = new int[n];
                for (int i = 0; i < m; i++) {
                    A[i] = in.nextInt();
                    B[i] = in.nextInt();
                    A[i]--;
                    B[i]--;
                    freq[A[i]]++;
                    freq[B[i]]++;
                }
                int threshold = (m + 1) / 2;
                boolean res = false;
                for (int i = 0; i < n; i++) {
                    if (freq[i] >= threshold && f(A, B, i)) {
                        res = true;
                    }
                }
                out.println(res ? "YES" : "NO");
            } catch (Exception e) {
            }
        }

        boolean f(int[] A, int[] B, int x) {
            int n = A.length, i = 0;
            for (; i < n; i++) {
                if (A[i] != x && B[i] != x) {
                    break;
                }
            }
            if (i == n) {
                return true;
            }
            int a = A[i], b = B[i];
            for (; i < n; i++) {
                if (A[i] != x && B[i] != x) {
                    if (a != A[i] && a != B[i]) {
                        a = -1;
                    }
                    if (b != A[i] && b != B[i]) {
                        b = -1;
                    }
                }
            }
            return a != -1 || b != -1;
        }

    }
}

