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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n, m; cin >> n >> m;
    vector<int> dg(n, 0);
    
    vector<vector<int>> g(n);

    for(int i = 0; i < m; i += 1) {
        int a, b; cin >> a >> b;
        a -= 1, b -= 1;
        dg[a] += 1;
        dg[b] += 1;
        g[a].push_back(b);
        g[b].push_back(a);
    }

    for(int i = 0; i < n; i += 1) {
        if(dg[i] % 2) {
            cout << "No\n";
            return 0;
        }
    }
    
    vector<int> fours;

    for(int i = 0; i < n; i += 1) {
        if(dg[i] >= 6) {
            cout << "Yes\n";
            return 0;
        } else if(dg[i] == 4) {
            fours.push_back(i);
        }
    }
    
    if(int(fours.size()) >= 3) {
        cout << "Yes\n";
        return 0;
    }

    if(int(fours.size()) < 2) {
        cout << "No\n";
        return 0;
    }

    int source = -1;
    for(int i = 0; i < n; i += 1)
        if(fours[0] != i) {
            source = i;
            break;
        }

    queue<int> q;
    q.push(source);
    vector<int> viz(n, 0);
    viz[source] = 1;

    while(not q.empty()) {
        int node = q.front();
        q.pop();

        for(auto v : g[node]) {
            if(v == fours[0])
                continue;
            if(not viz[v]) {
                viz[v] = 1;
                q.push(v);
            }
        }
    }

    if(accumulate(viz.begin(), viz.end(), 0) < n - 1) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
}