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;
const int N=2e5+10;
int n, m, dgr[N], S=0, T=0, cnt=0;
struct edge{
	int v, nxt;
}g[N<<1];
int tot=0, head[N];
inline void add(int u, int v){
	g[++tot]=edge{v,head[u]}; head[u]=tot;
	g[++tot]=edge{u,head[v]}; head[v]=tot;
}
inline void dfs(int u, int fa){
	if(u==S&&fa)return;
	if(u==T){
		++cnt; return;
	}
	for(int i=head[u]; i; i=g[i].nxt)if(g[i].v!=fa)dfs(g[i].v,u);
}
int main(){
	scanf("%d%d", &n, &m);
	if(m<n+2){
		puts("No"); return 0;
	}
	for(int i=1; i<=m; ++i){
		int u, v; scanf("%d%d", &u, &v);
		add(u,v); dgr[u]++; dgr[v]++;
	}
	for(int i=1; i<=n; ++i)
		if(dgr[i]&1){
			puts("No"); return 0;
		} else if(dgr[i]>=6){
			puts("Yes"); return 0;
		} else if(dgr[i]==4){
			if(!S)S=i;else if(!T)T=i;else{
				puts("Yes"); return 0;
			}
		}
	dfs(S,0);
	puts(cnt==4?"No":cnt==2?"Yes":"");
	return 0;
}
/*
6 8
1 3
3 2
1 4
4 2
1 5
5 2
1 6
6 2
*/