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;
#ifndef d
#define d(...)
#endif
#define st first
#define nd second
#define pb push_back
#define siz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
typedef long long LL;
typedef long double LD;
constexpr int INF=1e9+7;
constexpr LL INFL=1e18;
template<class L, class R> ostream &operator<<(ostream &os, pair<L,R> P) {
  return os << "(" << P.st << "," << P.nd << ")";
}
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template <typename T>
using ordered_set =
    tree<T, null_type, less<T>, rb_tree_tag,
         tree_order_statistics_node_update>;
*/

constexpr int maxn = 100005;

int n, m, deg[maxn];
vector<pair<int, int>> G[maxn];
bool used[maxn], on_stack[maxn];

vector<int> euler;

void dfs(int a) {
	while(not G[a].empty()) {
		int v, i;
		tie(v, i) = G[a].back();
		G[a].pop_back();
		if(used[i]) continue;
		used[i] = true;
		dfs(v);
		euler.push_back(a);
	}
}


int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	cin >> n >> m;
	for(int i=0; i<m; i++) {
		int a, b;
		cin >> a >> b;
		G[a].emplace_back(b, i);
		G[b].emplace_back(a, i);
		deg[a]++, deg[b]++;
	}
	
	for(int i=1; i<=n; i++)
		if(deg[i] % 2 == 1) {
			cout << "No\n";
			exit(0);
		}
	
	int root = mt19937(time(0))() % n + 1;
	dfs(root);
	
	stack<int> S;
	int cnt = 1;
	for(int x:euler) {
		if(on_stack[x]) {
			cnt++;
			while(on_stack[x]) {
				auto y = S.top();
				S.pop();
				on_stack[y] = false;
			}
		}
		on_stack[x] = true;
		S.push(x);
	}
	
	cout << (cnt >= 3 ? "Yes\n" : "No\n");	
}