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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=100005;
int n,m,i,j,k,d[N],head[N],adj[N*2],nxt[N*2];
bool v[N];
void dfs(int x)
{
	v[x]=true;
	for(int y=head[x];y;y=nxt[y])
		if(!v[adj[y]]&&d[adj[y]]<=2)
			dfs(adj[y]);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)
	{
		scanf("%d%d",&j,&k);
		++d[j],++d[k];
		adj[i*2-1]=k;
		nxt[i*2-1]=head[j];
		head[j]=i*2-1;
		adj[i*2]=j;
		nxt[i*2]=head[k];
		head[k]=i*2;
	}
	for(i=1,j=k=0;i<=n;++i)
	{
		if(d[i]&1)
		{
			printf("No");
			return 0;
		}
		if(d[i]>2)
			++j;
		if(d[i]>4)
			++k;
	}
	if(k)
	{
		printf("Yes");
		return 0;
	}
	if(j==0)
	{
		printf("No");
		return 0;
	}
	if(j>=3)
	{
		printf("Yes");
		return 0;
	}
	if(j==1)
	{
		printf("No");
		return 0;
	}
	for(i=1;i<=n;++i)
		if(d[i]>2)
			break;
	dfs(i);
	for(i=n;;--i)
		if(d[i]>2)
			break;
	for(j=head[i],k=0;j;j=nxt[j])
		if(v[adj[j]])
			++k;
	if(k==4)
		printf("No");
	else
		printf("Yes");
	return 0;
}