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 Python (3.4.3) (judge verdict: Wrong Answer)






#ルートを圧縮したい


N,M=input().split(" ")
N=int(N)
M=int(M)
path=[[]for i in range(N)]
for i in range(M):
	a,b=input().split(" ")
	path[int(a)-1].append(int(b)-1)
	path[int(b)-1].append(int(a)-1)



#圧縮
for i in range(len(path)):
	if(len(path[i])==2):
		path[path[i][0]].append(path[i][1])
		path[path[i][1]].append(path[i][0])
		path[path[i][0]].remove(i)
		path[path[i][1]].remove(i)
		path[i]=[]
odd=0
out=0
for i in path:
	out+=len(i)
	if(len(i)%2==1):
		odd=1

for i in range(len(path)):
	print(i,path[i])
print(out)
if(out>=6 and odd==0):
	print("Yes")
else:
	print("No")