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;
typedef long long ll;
typedef pair<ll,ll> P;
vector<P> node[200005];
bool used[200005]={false},used2[200005]={false},used3[200005]={false};
ll cnt[200005]={0};
bool dfs(ll pos,ll st,bool f){
  if(f&&pos==st) return true;
  for(int i=0;i<node[pos].size();i++){
    int to=node[pos][i].first,idx=node[pos][i].second;
    if(!used2[to]&&!used[idx]){
      used[idx]=true;
      used2[to]=true;
      if(dfs(to,st,true)) {
        used[idx]=true;
        return true;
      }
      else{
        used[idx]=false;
      }
    }
  }
  return false;
}
bool dfs2(ll pos,ll st,bool f){
  //cout<<pos<<" "<<st<<" "<<(f?"O":"X")<<endl;
  if(f&&pos==st) return true;
  for(int i=0;i<node[pos].size();i++){
    int to=node[pos][i].first,idx=node[pos][i].second;
    if(!used2[to]&&!used3[idx]&&!used[idx]){
      used3[idx]=true;
      used2[to]=true;
      if(dfs2(to,st,true)) {
        used3[idx]=true;
        return true;
      }
      else{
        if(!used[idx]) used3[idx]=false;
      }
    }
  }
  return false;
}
int main(){
  ll n,m;
  cin>>n>>m;
  for(int i=0;i<=max(n,m);i++){
    used[i]=false;
    used2[i]=false;
    used3[i]=false;
  }
  vector<P> qr;
  qr.push_back(P(0,0));
  for(int i=1;i<=m;i++){
    ll a,b;
    cin>>a>>b;
    qr.push_back(P(a,b));
    node[a].push_back(P(b,i));
    node[b].push_back(P(a,i));
    cnt[a]++;
    cnt[b]++;
  }
  ll cnt4=0;
  vector<ll> c4;
  for(int i=1;i<=n;i++){
    if(cnt[i]%2==1){
      cout<<"No"<<endl;
      return 0;
    }
    else if(cnt[i]>=6){
      cout<<"Yes"<<endl;
      return 0;
    }
    else if(cnt[i]==4) {
      cnt4++;
      c4.push_back(i);
    }
  }
  if(cnt4<2){
    cout<<"No"<<endl;
    return 0;
  }
  if(cnt4>2){
    cout<<"Yes"<<endl;
    return 0;
  }
  else if(cnt4==2){
    cout<<"No"<<endl;
    return 0;
    dfs(c4[0],c4[1],false);
    for(int i=1;i<=max(n,m);i++){
      used2[i]=false;
    }
    dfs2(c4[0],c4[1],false);
    for(int i=1;i<=max(n,m);i++){
      used2[i]=false;
      used[i]=used[i]|used3[i];
      used3[i]=false;
    }
    cout<<(dfs2(c4[0],c4[1],false)?"No":"Yes")<<endl;
  }
}