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: Runtime Error)

#include <bits/stdc++.h>
const int INF = 1e9;
const int MO = 1e9+7;
using LL = long long;
const LL LINF = 1e18;
const double EPS = 1e-10;
using namespace std;
#define COUT(v) cout<<(v)<<endl
#define CIN(n)  int(n);cin >> (n)
#define LCIN(n) LL(n);cin >> (n)
#define SCIN(n) string(n);cin >> (n)
#define YES(n) cout<<((n)? "YES" : "NO")<<endl
#define Yes(n) cout<<((n)? "Yes" : "No")<<endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) <<endl

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)

#define FOREACH(x,a) for(auto& (x) : (a) )

#define ALL(obj) (obj).begin(),(obj).end()

#define rt return
#define ENDL cout<<endl
#define MAXE(a,b) a=max(a,b)
#define MINE(a,b) a=min(a,b)
#define s second
#define f first

typedef string::const_iterator State;
class PalseError {};
class Edge{
public:
    LL from,to,value;
    Edge(LL a,LL b,LL c){
        from = a;
        to = b;
        value = c;
    }
    Edge(LL a,LL b){
        from = a;
        to = b;
    }
};

int main(){
    int N,M;
    cin >> N >> M;
    vector<int> zisuu(N);
    int cnt4 = 0;
    int cnt6 = 0;
    for(int a = 0;a < M;a++){
        int b,c;cin >> b >> c;
        b--,c--;
        zisuu.at(b)++;
        zisuu.at(c)++;
    }
    bool r = true;
    for(int a = 0;a < N;a++){
        if(zisuu.at(a)%2 == 1)r = false;
        if(zisuu.at(a) >= 4)cnt4++;
        if(zisuu.at(a) >= 6)cnt6++;
    }
    if(r && cnt4 >= 2 || cnt6 >= 1){
        cout<<"Yes"<<endl;
        return 1;
    }else{
        cout<<"No"<<endl;
    }
}