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: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int cnt;
int ans;
struct Edge{
	int v,nxt;
}e[200010];int tot=1;
int first[100010],deg[100010];
void build(int u,int v)
{
	e[++tot]=(Edge){v,first[u]};
	first[u]=tot;
	deg[u]++;
	return ;
}
bool book[100010];
bool flag;
void dfs1(int u)
{
	if(deg[u]!=2)flag=true;
	book[u]=true;
	for(int i=first[u];i;i=e[i].nxt)
		if(!book[e[i].v])dfs1(e[i].v);
	return ;
}

vector<int>to[100010];
bool vis[200010];
void link(int u,int fr)
{
	if(deg[u]==4){
		if(u==fr)ans++;
		else{
			to[u].push_back(fr);
			to[fr].push_back(u);
		}
		return ;
	}
	for(int& i=first[u];i;i=e[i].nxt)
		if(!vis[i])
		{
			vis[i]=vis[i^1]=true;
			link(e[i].v,fr);
		}
	return ;
}
bool book2[200010];
void dfs2(int u)
{
	book2[u]=true;
	for(int i=0;i<to[u].size();i++)
		if(!book2[to[u][i]])
			dfs2(to[u][i]);
	return ;
}


int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		build(u,v);
		build(v,u);
	}
	//check1
	for(int i=1;i<=n;i++)
		if(deg[i]&1){
			printf("No\n");
			return 0;
		}
	//check2
	for(int i=1;i<=n;i++)
		if(!book[i])
		{
			flag=false;
			cnt++;dfs1(i);
			ans+=1-flag;
		}
	if(cnt>3){
		printf("No\n");
		return 0;
	}
	//check3
	for(int i=1;i<=n;i++)
		if(deg[i]>=6){
			printf("Yes\n");
			return 0;
		}
	//link edges
	for(int i=1;i<=n;i++)
		if(deg[i]==4)
		{
			for(int j=first[i];j;j=e[j].nxt)
				if(!vis[j]){
					vis[j]=vis[j^1]=true;
					link(e[j].v,i);
				}
		}
	for(int i=1;i<=n;i++)
		if(!book2[i]&&deg[i]==4)
			dfs2(i),ans++;
	if(ans>=3)
		printf("Yes\n");
	else printf("No\n");
	return 0;
}