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 javafx.util.Pair;

import java.util.*;

public class Main {
    public static void main(String arg[]) {

        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt();
        int m=scanner.nextInt();
        int a[]=new int[m+1];
        int b[]=new int[m+1];
        BitSet bitSet1=new BitSet(m+1);
        BitSet bitSet2=new BitSet(m+1);
        int f=0;
        for(int i=1;i<=m;i++){
            a[i]=scanner.nextInt();
            b[i]=scanner.nextInt();
        }
        //a1
        for(int i=1;i<=m;i++){
            if(a[i]==a[1]||b[i]==a[1]){
                bitSet1.flip(i);bitSet2.flip(i);
            }
        }
        if(bitSet1.cardinality()==m)f=1;
        int id=bitSet1.nextClearBit(1);

        //ax
        for(int i=id;i<=m;i++){
            if(a[i]==a[id]||b[i]==a[id]){
                if(bitSet1.get(i)==false)
                    bitSet1.flip(i);
            }
        }if(bitSet1.cardinality()==m)f=1;

        //bx
        for(int i=id;i<=m;i++){
            if(a[i]==b[id]||b[i]==b[id]){
                if(bitSet2.get(i)==false)
                    bitSet2.flip(i);
            }
        }if(bitSet2.cardinality()==m)f=1;

        bitSet1.clear();bitSet2.clear();

        //b1
        for(int i=1;i<=m;i++){
            if(a[i]==b[1]||b[i]==b[1]){
                bitSet1.flip(i);bitSet2.flip(i);
            }
        }
        if(bitSet1.cardinality()==m)f=1;
        id=bitSet1.nextClearBit(1);

        //ax
        for(int i=id;i<=m;i++){
            if(a[i]==a[id]||b[i]==a[id]){
                if(bitSet1.get(i)==false)
                    bitSet1.flip(i);
            }
        }if(bitSet1.cardinality()==m)f=1;

        //bx
        for(int i=id;i<=m;i++){
            if(a[i]==b[id]||b[i]==b[id]){
                if(bitSet2.get(i)==false)
                    bitSet2.flip(i);
            }
        }if(bitSet2.cardinality()==m)f=1;

        if(f==1)System.out.println("YES");
        else System.out.println("NO");

    }


}
