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)

#include <bits/stdc++.h>
using namespace std;
// #define int long long
// #define double long double
#define rep(i,n) for (int i=0; i<(int)(n); ++i)
#define rep1(i,n) for (int i=1; i<(int)(n); ++i)
#define repeq(i,n) for (int i=0; i<=(int)(n); ++i)
#define rep1eq(i,n) for (int i=1; i<=(int)(n); ++i)
#define rrep(i,n) for (int i=(int)(n)-1; i>=0; --i)
#define rrep1(i,n) for (int i=(int)(n)-1; i>0; --i)
#define rrepeq(i,n) for (int i=(int)(n); i>=0; --i)
#define rrep1eq(i,n) for (int i=(int)(n); i>0; --i)
#define REP(i,a,b) for (int i=(int)(a); i<=(int)(b); ++i)
#define RREP(i,a,b) for (int i=(int)(a); i>=(int)(b); --i)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
template<typename T> using Graph = vector<vector<T>>;
template<typename T> using Spacial = vector<vector<vector<T>>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename T> using greater_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const int MOD = 1e9+7;
const int MOD2 = 998244353;
// const double EPS = 1e-9;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
string interval[2] = {" ", "\n"}; // {" ", "\n"}

template<typename T> struct is_plural : false_type{};
template<typename T1, typename T2> struct is_plural<pair<T1, T2>> : true_type{};
template<typename T> struct is_plural<vector<T>> : true_type{};
template<typename T> struct is_plural<complex<T>> : true_type{};
 
template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << " " << p.second; }
template<typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto itr = vec.begin(); itr != vec.end(); ++itr) is >> *itr; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { if (vec.empty()) return os; bool pl = is_plural<T>(); os << vec.front(); for (auto itr = ++vec.begin(); itr != vec.end(); ++itr) os << interval[pl] << *itr; return os; }
 
bool CoutYN(bool a, string y = "Yes", string n = "No") { cout << (a ? y : n) << "\n"; return a; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

long long modpow(int a, long long n, int mod = MOD) { long long ret = 1; do { if (n & 1) ret = ret * a % mod; a = 1LL * a * a % mod; } while (n >>= 1); return ret; }

template<typename T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
template<typename T> T LCM(T a, T b) { return a / GCD(a, b) * b; }

template<typename T1, typename T2> bool CompareBySecond(pair<T1, T2> a, pair<T1, T2> b) { return a.second != b.second ? a.second < b.second : a.first < b.first; }
template<typename T1, typename T2> bool CompareByInverse(pair<T1, T2> a, pair<T1, T2> b) { return a.first != b.first ? a.first < b.first : a.second > b.second; }


/* -------- <templates end> -------- */


void solve() {
  int n, m; cin >> n >> m;
  vector<set<int>> g(n);
  rep(_,m) {
    int a, b; cin >> a >> b; --a, --b;
    g[a].emplace(b);
    g[b].emplace(a);
  }

  rep(i,n) {
    if (g[i].size() & 1) {
      cout << "No\n";
      return;
    }
  }

  auto detection = [&]() -> bool {
    vb seen(n, false);
    stack<int> cycle;

    auto dfs = [&](auto dfs, int v, int p) -> bool {
      if (seen[v]) {
        int a, b = v;
        while (1) {
          a = b, b = cycle.top(); cycle.pop();
          g[a].erase(b);
          g[b].erase(a);
          --m;
          if (b == v) break;
        }
        return true;
      }

      seen[v] = true;
      cycle.emplace(v);

      for (auto nv : g[v]) {
        if (nv != p) {
          if (dfs(dfs, nv, v)) return true;
        }
      }
      return false;
    };

    rep(i,n) {
      if (!g[i].empty()) return dfs(dfs, i, -1) && m;
    }
    return false;
  };

  CoutYN(detection() && detection());
}


/* -------- <programs end> -------- */


signed main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  solve();
  return 0;
}