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>
#define debug(...) fprintf(stderr, __VA_ARGS__)

using namespace std;
const int N = 400005;

inline int input() {
	int x; char ch; while (!isdigit(ch = getchar()));
	for (x = ch ^ 48; isdigit(ch = getchar()); x = x * 10 + (ch ^ 48));
	return x;
}

inline void chkmin(int &x, int y) { x = x < y ? x : y; }

int n, m;
int a[N], b[N], du[N];
int head[N], ne[N], to[N], cnte = 1;
int ans[N], tot, stk[N], top;
bool vis[N], ins[N];

inline void AddEdge(int x, int y) {
	ne[++cnte] = head[x], head[x] = cnte, to[cnte] = y;
	ne[++cnte] = head[y], head[y] = cnte, to[cnte] = x;
}

void dfs(int u) {
	for (int i = head[u]; i; i = ne[i]) {
		if (vis[i]) continue;
		vis[i] = vis[i ^ 1] = true;
		dfs(to[i]);
	}
	ans[++tot] = u;
}

bool check() {
	int cnt = 0;
	dfs(1), reverse(ans + 1, ans + tot + 1);
//	for (int i = 1; i <= tot; ++i)
//		debug("%d\n", ans[i]);
	for (int i = 1; i <= tot; ++i) {
		if (ins[ans[i]]) {
			while (stk[top] != ans[i])
				ins[stk[top--]] = false;
			++cnt;
		} else {
			stk[++top] = ans[i];
			ins[ans[i]] = true;
		}
	}
	return cnt >= 3;
}

int main() {
#ifdef local
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
//	freopen("log", "w", stderr);
#endif
	ios::sync_with_stdio(false);

	cin >> n >> m;

	for (int i = 1; i <= m; ++i) {
		cin >> a[i] >> b[i];
		++du[a[i]], ++du[b[i]];
		AddEdge(a[i], b[i]);
	}

	for (int i = 1; i <= n; ++i)
		if (du[i] & 1) return puts("No"), 0;
	if (m - n >= 3) puts("Yes");
	else if (m - n == 2) puts(check() ? "Yes" : "No");
	else puts("No");
	return 0;
}