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 Main {
  static InputStream inputStream = System.in;
  static OutputStream outputStream = System.out;
  static InputReader in = new InputReader(inputStream);
  static PrintWriter out = new PrintWriter(outputStream);
  
  public static void main(String[] args) {
    int n = in.nextInt();
    int m = in.nextInt();

    int[] qtyX = new int[n + 1];
    int[] qtyY = new int[n + 1];
    int withoutX = 0;
    int withoutY = 0;

    int x = in.nextInt();
    int y = in.nextInt();

    for (int i = 0; i < m - 1; i++) {
      int a = in.nextInt();
      int b = in.nextInt();

      if (a != x && b != x) {
        qtyX[a]++;
        qtyX[b]++;
        withoutX++;
      }
      if (a != y && b != y) {
        qtyY[a]++;
        qtyY[b]++;
        withoutY++;
      }
    }

    boolean OK = false;
    for (int i = 1; i < n + 1 && !OK; i++) {
      if (qtyX[i] == withoutX || qtyY[i] == withoutY) {
        OK = true;
      }
    }

    out.println(OK ? "YES" : "NO");


    out.close();
  }

  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream), 32768);
      tokenizer = null;
    }

    public String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

    public long nextLong () {
      return Long.parseLong(next());
    }

    public char nextChar () {
      return next().charAt(0);
    }
  }
}