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 Ruby (2.3.3) (judge verdict: Wrong Answer)

class UnionFind
	def initialize(n); @p = *0...n; @r = [0] * n; end; def find(x); @p[x] == x ? x : @p[x] = find(@p[x]); end; def unite(x, y); x = find(x); y = find(y); return if x == y; if @r[x] < @r[y]; @p[x] = y; else; @p[y] = x; @r[x] += 1 if @r[x] == @r[y]; end; nil; end; def same(x, y); find(x) == find(y); end

	def par
		@p.map{|i|find(i)}
	end
end


(n,m),*ab=$<.map{|s|s.split.map &:to_i}
v=Array.new(n){[]}
uf=UnionFind.new(n)
ab.each{|a,b|
a-=1
b-=1
v[a]<<b
v[b]<<a
uf.unite(a,b)
}
v.delete_if{|s|s.size==0}
if v.any?{|s|s.size%2==1}
	puts "No"
else
	s=v.map{|x|x.size/2-1}.reduce(:+)
	if s>=2
		puts "Yes"
	else
		k=(uf.par-[0]).uniq.size
		if k+s>=3
			puts "Yes"
		else
			puts "No"
		end
	end
end