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;
typedef long long ll;

int main(){

    int N, M;
    cin >> N >> M;
    vector<int>vec(N);
    for(int i = 0; i < M; i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        vec[a]++;
        vec[b]++;
    }
    sort(vec.begin(), vec.end(), greater<int>());
    int fail = 0;
    for(int i = 0; i < N; i++){
        if(vec[i] % 2)fail = 1;
    }
    if(fail){
        cout << "No" << endl;
        return 0;
    }
    if(vec[0]>=6){
        cout << "Yes" << endl;
        return 0;
    }
    if(N>1){
        if(vec[1]>=4){
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;

    return 0;
}