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;
#define rep(i,a,b) for (int i = (a); i <= (b); ++ i)
#define rrp(i,a,b) for (int i = (a); i >= (b); -- i)
#define gc() getchar()
#define fir first
#define sec second
typedef pair<int,int> pii;
typedef long double db;
typedef long long ll;
template <typename tp>
inline void read(tp& x) {
  x = 0; char tmp; bool key = 0;
  for (tmp = gc(); !isdigit(tmp); tmp = gc())
    key = (tmp == '-');
  for (; isdigit(tmp); tmp = gc()) 
    x = (x << 3) + (x << 1) + (tmp ^ '0');
  if (key) x = -x;
}
template <typename tp>
inline void ckmn(tp& x,tp y) {
  x = x < y ? x : y;
}
template <typename tp>
inline void ckmx(tp& x,tp y) {
  x = x < y ? y : x;
}

const int N = 100010;
struct edge {
  int la,b;
} con[N << 1];
int tot, fir[N];
void add_edge(int from,int to) {
  con[++tot] = (edge) {fir[from], to};
  fir[from] = tot;
}
int n, m, deg[N], en, st, num;
void dfs(int pos,int fa) {
  if (pos == en) {
    ++ num;
    return;
  }
  for (int i = fir[pos]; i; i = con[i].la) {
    if (con[i].b == fa) continue;
    if (con[i].b == st) continue;
    dfs(con[i].b, pos);
  }
}
int main() {
  int x,y;
  read(n), read(m);
  rep (i, 1, m) {
    read(x), read(y);
    add_edge(x, y);
    add_edge(y, x);
    ++ deg[x], ++ deg[y];
  }
  rep (i, 1, n) if (deg[i] % 2 == 1) {
    puts("No");
    return 0;
  }
  rep (i, 1, n) if (deg[i] >= 6) {
    puts("Yes");
    return 0;
  }
  int num = 0;
  rep (i, 1, n) if (deg[i] >= 4) ++ num;
  if (num >= 3) return puts("Yes"), 0;
  if (num <= 1) return puts("No"), 0;
  int px = 0, py = 0;
  rep (i, 1, n) if (deg[i] == 4) {
    if (!px) px = i;
    else py = i;
  }
  en = py;
  st = px;
  dfs(px, 0);
  if (num == 4) puts("Yes");
  else puts("No");
  return 0;
}