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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef double D;
typedef long double LD;
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> opset;
#define SZ(X) ((int)X.size())
#define ALL(I) I.begin(), I.end()
#define PB push_back
#define F first
#define S second
#define MP make_pair
#define eps 1e-12
#define INF 1000000007
#define MXN 200005
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)

vector<int> v[MXN];
int n, m, fr, to;
int vis[MXN], bk[MXN], dep[MXN];

void dfs(int x, int pa){
	dep[x] = dep[pa] + 1;
	for(auto i : v[x]){
		if(i == pa) continue;
		if(vis[i]){
			if(dep[i] < dep[x]) bk[i] ++;
		}
		else {
			vis[x] ++;
			dfs(i, x);
		}
	}
}

int main(){
	scanf("%d%d", &n, &m);
	while(m --){
		scanf("%d%d", &fr, &to);
		v[fr].PB(to);
		v[to].PB(fr);
	}
	for(int i = 1; i <= n; i ++){
		//if(SZ(v[i]) & 1){
			puts("No");
			return 0;
		//}
	}
	dfs(1, 1);
	int cnt = 0;
	for(int i = 1; i <= n; i ++){
		if(bk[i]) cnt += (vis[i] + bk[i]) >> 1;
	}
	if(cnt < 3) puts("No");
	else puts("Yes");
}