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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCorder
{
	public class Program
	{
		static void Main(string[] args)
		{
			int[] nm = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
			int n = nm[0];
			int m = nm[1];
			List<int>[] to = new List<int>[n];
			for (int i = 0; i < n; i++) {
				to[i] = new List<int>();
			}
			for (int i = 0; i < m; i++) {
				int[] ab = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
				to[ab[0] - 1].Add(ab[1] - 1);
				to[ab[1] - 1].Add(ab[0] - 1);
			}

			int cross6 = 0;
			List<int> cross = new List<int>();
			for (int i = 0; i < n; i++) {
				if (to[i].Count % 2 == 1) {
					Console.WriteLine("No");
					return;
				}

				if (to[i].Count != 2) {
					cross.Add(i);
					if (to[i].Count >= 6) {
						++cross6;
					}
				}
			}

			if (cross.Count >= 2 || cross6 > 0) {
				if (cross.Count == 2 && cross6 == 0) {
					int same = 0;
					foreach (int v in to[cross[0]]) {
						if (to[cross[1]].Contains(v)) {
							same++;
						}
					}

					if (same > 2) {
						Console.WriteLine("No");
					} else {
						Console.WriteLine("Yes");
					}
				} else {
					Console.WriteLine("Yes");
				}
			} else {
				Console.WriteLine("No");
			}
		}
	}
}