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 9.2.1) (judge verdict: Wrong Answer)

#line 2 "/Users/kaage/Desktop/ProgrammingWorkspace/library/other/template.hpp"
#define _CRT_SECURE_NO_WARNINGS
#pragma target("avx2")
#pragma optimize("O3")
#pragma optimize("unroll-loops")
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define all(V) V.begin(),V.end()
typedef unsigned int uint;
typedef long long lint;
typedef unsigned long long ulint;
typedef std::pair<int, int> P;
typedef std::pair<lint, lint> LP;
constexpr int INF = INT_MAX/2;
constexpr lint LINF = LLONG_MAX/2;
constexpr double eps = DBL_EPSILON;
constexpr double PI=3.141592653589793238462643383279;
template<class T>
class prique :public std::priority_queue<T, std::vector<T>, std::greater<T>> {};
template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
	if (lhs < rhs) {
		lhs = rhs;
		return 1;
	}
	return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
	if (lhs > rhs) {
		lhs = rhs;
		return 1;
	}
	return 0;
}
inline lint gcd(lint a, lint b) {
	while (b) {
		lint c = a;
		a = b; b = c % b;
	}
	return a;
}
inline lint lcm(lint a, lint b) {
	return a / gcd(a, b) * b;
}
bool isprime(lint n) {
	if (n == 1)return false;
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0)return false;
	}
	return true;
}
template<typename T>
T mypow(T a, lint b) {
	T res(1);
	while(b){
		if(b&1)res*=a;
		a*=a;
		b>>=1;
	}
	return res;
}
lint modpow(lint a, lint b, lint m) {
	lint res(1);
	while(b){
		if(b&1){
			res*=a;res%=m;
		}
		a*=a;a%=m;
		b>>=1;
	}
	return res;
}
template<typename T>
void printArray(std::vector<T>& vec) {
	rep(i, vec.size()){
		std::cout << vec[i];
		std::cout<<(i==(int)vec.size()-1?"\n":" ");
	}
}
template<typename T>
void printArray(T l, T r) {
	T rprev = std::prev(r);
	for (T i = l; i != rprev; i++) {
		std::cout << *i << " ";
	}
	std::cout << *rprev << std::endl;
}
LP extGcd(lint a,lint b) {
	if(b==0)return {1,0};
	LP s=extGcd(b,a%b);
	std::swap(s.first,s.second);
	s.second-=a/b*s.first;
	return s;
}
LP ChineseRem(const lint& b1,const lint& m1,const lint& b2,const lint& m2) {
	lint p=extGcd(m1,m2).first;
	lint tmp=(b2-b1)*p%m2;
	lint r=(b1+m1*tmp+m1*m2)%(m1*m2);
	return std::make_pair(r,m1*m2);
}
template<typename F>
inline constexpr decltype(auto) lambda_fix(F&& f){
	return [f=std::forward<F>(f)](auto&&... args){
		return f(f,std::forward<decltype(args)>(args)...);
	};
}
#line 2 "main.cpp"
int N,M,count[100010];
int main(){
	std::cin>>N>>M;
	rep(i,M){
		int a,b;
		std::cin>>a>>b;
		count[a]++;count[b]++;
	}
	REP(i,N){
		if(count[i]&1){
			puts("No");
			return 0;
		}
	}
	std::sort(count+1,count+N+1);
	if(count[N]>=4&&N-1>0&&count[N-1]<=count[N])puts("Yes");
	else puts("No");
	return 0;
}