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: Runtime Error)

#include <algorithm>
#include <cassert>
#include <cctype>
#include <chrono>
#define _USE_MATH_DEFINES
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <tuple>
#include <vector>
using namespace std;

#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()

const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int MOD = 1000000007; // 998244353;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
/*-------------------------------------------------*/
int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  // freopen("input.txt", "r", stdin);

  int n, m; cin >> n >> m;
  vector<int> cnt(n, 0);
  REP(i, m) {
    int a, b; cin >> a >> b; --a; --b;
    ++cnt[a];
    ++cnt[b];
  }
  REP(i, n) {
    if (cnt[i] & 1) {
      cout << "No\n";
      return 0;
    }
  }
  assert(false);
  cout << (m - n >= 2 ? "Yes" : "No") << '\n';
  return 0;
}