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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define N_MAX 131072
#define A(I)  (e[(I)]/N_MAX)
#define B(I)  (e[(I)]%N_MAX)
#define FORN  for(i=1;i<=N;i++)
void Sort(int l,int r);
char* ans(void);
long e[200001];
int main(void){
  printf("%s",ans());
  return 0;
}
char* ans(void){
  int N,M,w[100001]={},index[100001],i,j,v4[2],n_v4=0,t,t2;
  scanf("%d %d",&N,&M);
  for(i=1;i<=M;i++){
    scanf("%d %d",&t,&t2);
    e[i]=(long)t*N_MAX+t2;
    e[i+M]=(long)t2*N_MAX+t;
    w[t]++;
    w[t]++;    
  }
  FORN{if(w[i]%2==1){return "No";}}
  FORN{if(w[i]>=6){return "Yes";}}
  FORN{if(w[i]==4){v4[n_v4++]=i;}}
  if(n_v4>=3){return "Yes";}
  else if(n_v4==2){
    Sort(1,2*M);
    j=1;
    FORN{
      while(A(j)!=i){j++;}
      index[i]=j;
    }
    for(i=0;i<3;i++){
      t=B(index[v4[0]]+i);
      t2=v4[0];
      while((t!=v4[0])&&(t!=v4[1])){
        if(B(index[t])==t2){
          t2=t;
          t=B(index[t]+1);
        }
        else{
          t2=t;
          t=B(index[t]);
        }
      }
      if(t==v4[0]){return "Yes";}
    }
  }
  return "No";
}

void Sort(int l,int r)
{
  int i,j;
  long sta,temp;
  i=l;
  j=r;
  sta=e[(l+r)/2];
  while(1){
    while(e[i]<sta){i++;}
    while(sta<e[j]){j--;}
    if(i>=j){break;}
    temp=e[i];
    e[i]=e[j];
    e[j]=temp;
    i++;
    j--;
  }
  if (l<i-1){Sort(l,i-1);} 
  if (j+1<r){Sort(j+1,r);}
}