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 <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int maxn = 100100;
struct edges
{
	int to, next;
}edge[maxn * 2];
int point[maxn] = {0}, te = 1;
void add(int a, int b)
{
	te ++;
	edge[te].to = b;
	edge[te].next = point[a];
	point[a] = te;
}
bool flag[maxn * 2] = {0};
int du[maxn] = {0};
bool dfs(int now, int ed, int p)
{
	if(p && now == ed)
		return 1;
	du[now] --;
	//cout << now << " " << ed << " " << p << endl;
	for(int e = point[now]; e; e = edge[e].next)
		if(!flag[e])
		{
			//cout << edge[e].to << endl;
			du[edge[e].to] --;
			flag[e] = flag[e ^ 1] = 1;
			return dfs(edge[e].to, ed, now);
		}
	return 0;
}
int n, m;
int main()
{
	std::ios::sync_with_stdio(false);
	cin >> n >> m;
	for(int i = 1; i <= m; i ++)
	{
		int a, b;
		cin >> a >> b;
		du[a] ++;
		du[b] ++;
		add(a, b);
		add(b, a);
	}
	int st[3], cnt = 0;
	bool asd = 0;
	for(int i = 1; i <= n; i ++)
	{
		if(du[i] % 2 == 1)
		{
			cout << "No" << endl;
			return 0;
		}
		if(du[i] >= 6)
		{
			asd = 1;
			continue;
		}
		if(du[i] > 2)
		{
			cnt ++;
			if(cnt > 2)
			{
				asd = 1;
				continue;
			}
			st[cnt] = i;
		}
	}
	if(asd)
	{
		cout << "Yes" << endl;
		return 0;
	}
	if(cnt < 2)
	{
		cout << "No" << endl;
		return 0;
	}
	dfs(st[1], st[2], 0);
	dfs(st[1], st[2], 0);
	if(dfs(st[1], st[2], 0))
	{
		cout << "No" << endl;
		return 0;
	}
	/*
	for(int i = 1; i <= n; i ++)
		cout << du[i] << " " ;
	cout << endl;
	*/
	cout << "Yes" << endl;
	//cerr << "Yes" << endl;
	return 0;
}
/*
4 6
1 2
2 3
1 3
1 3
1 4
4 3
*/
/*
10 14
7 1
1 3
3 9
9 1
1 4
4 6
6 2
4 8
8 10
10 8
8 5
5 6
6 1
1 2
*/