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;
using ll=long long;
using vb=vector<bool>;
using vvb=vector<vb>;
using vd=vector<double>;
using vvd=vector<vd>;
using vi=vector<int>;
using vvi=vector<vi>;
using vl=vector<ll>;
using vvl=vector<vl>;
using pii=pair<int,int>;
using pll=pair<ll,ll>;
using tll=tuple<ll,ll>;
using tlll=tuple<ll,ll,ll>;
using vs=vector<string>;

#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rep(i,n) range(i,0,n)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define range(i,a,n) for(int i=(a);i<(n);i++)

#define LINF ((ll)1ll<<60)
#define INF ((int)1<<30)
#define EPS (1e-9)
#define MOD (1000000007ll)
#define fcout(a) cout<<setprecision(a)<<fixed
#define fs first
#define sc second
#define PI 3.1415926535897932384

int dx[]={1,0,-1,0,1,-1,-1,1},dy[]={0,1,0,-1,1,1,-1,-1};

template<class T>bool chmax(T&a,T b){if(a<b){a=b; return true;}return false;}
template<class T>bool chmin(T&a,T b){if(a>b){a=b; return true;}return false;}
template<class S>S acm(vector<S>&a){return accumulate(all(a),S());}
template<class S>S max(vector<S>&a){return *max_element(all(a));}
template<class S>S min(vector<S>&a){return *min_element(all(a));}
void YN(bool b){cout<<(b?"YES":"NO")<<"\n";}
void Yn(bool b){cout<<(b?"Yes":"No")<<"\n";}
void yn(bool b){cout<<(b?"yes":"no")<<"\n";}

ll max(int a,ll b){return max((ll)a,b);} ll max(ll a,int b){return max(a,(ll)b);}
template<class T>void puta(T&&t){cout<<t<<"\n";}
template<class H,class...T>void puta(H&&h,T&&...t){cout<<h<<' ';puta(t...);}
template<class S,class T>ostream&operator<<(ostream&os,pair<S,T>p){os<<"["<<p.first<<", "<<p.second<<"]";return os;};
template<class S>auto&operator<<(ostream&os,vector<S>t){bool a=1; for(auto s:t){os<<(a?"":" ")<<s;a=0;} return os;}
#define _cTime (chrono::system_clock::now())
#define progress (chrono::duration_cast<chrono::milliseconds>(_cTime-_sTime).count())
#define reset _sTime=_cTime
auto reset;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n,m;
	cin>>n>>m;
	vi v(n);
	int a,b;
	rep(i,m){
		cin>>a>>b;
		a--;b--;
		v[a]++;v[b]++;
	}
	bool ans=true;
	bool six=false;
	rep(i,n)ans&=(v[i]%2==0),six|=v[i]>=6;
	Yn(ans&&six);
}