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<bits/stdc++.h>
using namespace std;
 
template <typename T> void chmin(T&x,const T &y)
{
	if(x>y)x=y;
}
template <typename T> void chmax(T &x,const T &y)
{
	if(x<y)x=y;
}
typedef long long s64;
typedef unsigned long long u64;
typedef unsigned int u32;
typedef pair<int,int> pii;
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define per(i,r,l) for(int i=r;i>=l;--i)
#define rep0(i,l,r) for(int i=l;i<r;++i)
#define gc (c=getchar())
int read()
{
	char c;
	while(gc<'-');
	if(c=='-')
	{
		int x=gc-'0';
		while(gc>='0')x=x*10+c-'0';
		return -x;
	}
	int x=c-'0';
	while(gc>='0')x=x*10+c-'0';
	return x;
}
#undef gc
 
const int N=1e5+5;
vector<int>lk[N];
bool vis[N];
 
int dfs(int x,int fr)
{
	if(vis[x]) return 0;
	if(lk[x].size()>2)return x;
	vis[x]=1;
	return dfs(lk[x][lk[x][0]==fr],x);
}
 
int main()
{
	int n,m;
	cin>>n>>m;
	if(m<n+2){puts("No");exit(0);}
	while(m--)
	{
		int x=read(),y=read();
		lk[x].push_back(y);
		lk[y].push_back(x);
	}
	rep(i,1,n)
	if(lk[i].size()%2){puts("No");exit(0);}
	if(m>n+2){puts("Yes");exit(0);}
	rep(i,1,n)
	if(lk[i].size()>4){puts("Yes");exit(0);}
	//cout<<"y"<<endl;
	int cnt=0;
	rep(i,1,n)
	if(lk[i].size()==4)
	{
		for(auto j:lk[i])
		{
			if(dfs(j,i)!=i)++cnt;
		}
		break;
	}
	if(cnt>=4)puts("No");
	else puts("Yes");	
}
/*
7 9
1 2
2 3
1 3
3 4
3 5
4 5
5 6
5 7
6 7
*/