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<cstdio>
using namespace std;
int n,m,cnt,sz,ans;
int fir[101010],nxt[202020],to[202020],q[101010],d[101010];
int du[101010],st[101010],top,bok[202020];
char S[10][10]={"No","Yes"};
void add(int x,int y)
{
	nxt[++cnt]=fir[x];fir[x]=cnt;
	to[cnt]=y;
}
void dfs(int x,int fa)
{
	for(int i=fir[x];i;i=nxt[i])
	if(to[i]!=fa&&!bok[(i+1)/2])
	{
		fir[x]=nxt[fir[x]];
		bok[(i+1)/2]=1;
		dfs(to[i],x);
	}
	else fir[x]=nxt[fir[x]];
	q[++sz]=x;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
		du[x]++;du[y]++;
	}
	for(int i=1;i<=n;i++)
	if(du[i]&1)
	{
		puts("No");
		return 0;
	}
	else if(du[i]>4)
	{
		puts("Yes");
		return 0;
	}
	dfs(1,0);
//	for(int i=1;i<=sz;i++)
//	cout<<q[i]<<" ";
//	puts("");
	for(int i=1;i<=sz;i++)
	if(!d[q[i]])
	{
		st[++top]=q[i];
		d[q[i]]=1;
	}
	else
	{
		while(st[top]!=q[i]) d[st[top--]]=0;
		ans++;
	}
	printf("%s\n",S[ans>=3]);
	return 0;
}