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;



class Solution {
  int n, m;
  vector<int> deg;
public:
  void run() {
    cin >> n >> m;
    deg.resize(n);
    for (int i = 0; i < m; ++i) {
      int a, b; cin >> a >> b;
      --a, --b;
      ++deg[a], ++deg[b];
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (deg[i] % 2) {
        cout << "No\n";
        exit(0);
      }
      ans += deg[i] / 2 - 1;
    }
    cout << (ans > 1 ? "Yes\n" : "No\n");
  }
};


int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);
  Solution().run();
}