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>
#define maxn 100050
using namespace std;
int n,m;
int deg[maxn];
vector<int> G[maxn];
int fa[maxn];
bool flag,vis[maxn];
set<int> aban[maxn];
bool dfs(int i)    {
    vis[i]=1;
    for (int j:G[i]) {
        if (fa[i]==j||aban[i].count(j))   continue;
        if (!vis[j])
            fa[j]=i,dfs(j);
        else
        if (!flag)  {
            flag=1;
            aban[i].insert(j);
            aban[j].insert(i);
            for (int k=i;k!=fa[j];k=fa[k])  {
                // cout<<"del:"<<k<<endl;
                deg[k]-=2;
                if (k==j)   continue;
                aban[k].insert(fa[k]);
                aban[fa[k]].insert(k);
            }
        }
    }
}
int main()  {
    scanf("%d%d",&n,&m);
    while (m--) {
        int u,v;
        scanf("%d%d",&u,&v);
        G[u].push_back(v),G[v].push_back(u);
        ++deg[u],++deg[v];
    }
    for (int i=1;i<=n;++i)  {
        if (deg[i]&1) 
            return puts("No"),0;
    }
    for (int c=0;c<3;++c)   {
        memset(vis,0,sizeof(vis));
        flag=0;
        for (int i=1;i<=n;++i)
            if (deg[i]) {
                dfs(i);
                break;
            }
        if (!flag) return puts("No"),0;
    }

    puts("Yes");
    return 0;
}