Problem:
Score : 800 points

Problem Statement

You are given a simple connected undirected graph consisting of N vertices and M edges.
The vertices are numbered 1 to N, and the edges are numbered 1 to M.

Edge i connects Vertex a_i and b_i bidirectionally.

Determine if three circuits (see Notes) can be formed using each of the edges exactly once.

Notes

A circuit is a cycle allowing repetitions of vertices but not edges.

Constraints

All values in input are integers.

1 \leq N,M \leq 10^{5}

1 \leq a_i, b_i \leq N

The given graph is simple and connected.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_M b_M

Output

If three circuits can be formed using each of the edges exactly once, print Yes; if they cannot, print No.

Sample Input 1

7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7

Sample Output 1

Yes

Three circuits can be formed using each of the edges exactly once, as follows:

Sample Input 2

3 3
1 2
2 3
3 1

Sample Output 2

No

Three circuits are needed.

Sample Input 3

18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15

Sample Output 3

Yes

Solution in Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import static java.lang.System.*;
import java.util.*;

class UnionFind {
    int[] par;
    int[] num;

    UnionFind(int n) { 
    	par = new int[n];
    	num = new int[n];
        for(int i = 0; i < n; i++) {
        	par[i] = i;
        	num[i] = 1;
        }
    }

    int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) {
        int rx = root(x); 
        int ry = root(y);
        if (rx == ry) return; 

        par[rx] = ry;
        num[ry] += num[rx];
    }

    boolean same(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }

    int getNum(int n) {
    	return num[root(n)];
    }
}

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(in);

        int n = sc.nextInt();
		int m = sc.nextInt();
		
		int[] count = new int[n];
		ArrayList<ArrayList<Integer>> sides = new ArrayList<ArrayList<Integer>>();
		for(int i=0; i<n; i++) sides.add(new ArrayList<Integer>());
		for(int i=0; i<m; i++) {
			int a = sc.nextInt()-1;
			int b = sc.nextInt()-1;
			
			count[a]++;
			count[b]++;
			sides.get(a).add(b);			
		}
		
		boolean even = true;
		boolean six = false;
		int branch = 0;
		for(int i=0; i<n; i++) {
			even &= ((count[i]%2)==0);
			six |= count[i] >= 6;
			branch += count[i]-2;
		}
		if(branch >= 6) six = true;

		if(even && !six && branch == 4) {
			UnionFind uf = new UnionFind(n);
			for(int i=0; i<n; i++) {
				if(sides.get(i).size() == 2 && sides.get(sides.get(i).get(0)).size() == 2 && 
				                    	       sides.get(sides.get(i).get(1)).size() == 2) {
					uf.unite(i, sides.get(i).get(0));
					uf.unite(i, sides.get(i).get(1));
				}
			}
			
			ArrayList<Integer> mid = new ArrayList<Integer>();
			for(int i=0; i<n; i++) {
				if(sides.get(i).size() == 4) {
					if(uf.same(sides.get(i).get(0), sides.get(i).get(1))) {
						mid.add(sides.get(i).get(2));
						mid.add(sides.get(i).get(3));
					}
					else if(uf.same(sides.get(i).get(0), sides.get(i).get(2))) {
						mid.add(sides.get(i).get(1));
						mid.add(sides.get(i).get(3));
					}						
					else if(uf.same(sides.get(i).get(0), sides.get(i).get(3))) {
						mid.add(sides.get(i).get(1));
						mid.add(sides.get(i).get(2));
					}						
					else if(uf.same(sides.get(i).get(1), sides.get(i).get(2))) {
						mid.add(sides.get(i).get(0));
						mid.add(sides.get(i).get(3));
					}						
					else if(uf.same(sides.get(i).get(1), sides.get(i).get(3))) {
						mid.add(sides.get(i).get(0));
						mid.add(sides.get(i).get(2));
					}
					else if(uf.same(sides.get(i).get(2), sides.get(i).get(3))) {
						mid.add(sides.get(i).get(0));
						mid.add(sides.get(i).get(1));
					}						
				}	
			}
			if(mid.size() == 4) {
				if(uf.same(mid.get(0), mid.get(2)) && uf.same(mid.get(1), mid.get(3)) || 
			       uf.same(mid.get(0), mid.get(3)) && uf.same(mid.get(1), mid.get(2))) 
					six = true;
			}
		}
		
		out.println(even && six ? "Yes": "No");
    }
}