Problem:
Score : 500 points

Problem Statement

Rng has a connected undirected graph with N vertices.
Currently, there are M edges in the graph, and the i-th edge connects Vertices A_i and B_i.

Rng will add new edges to the graph by repeating the following operation:

Operation: Choose u and v (u \neq v) such that Vertex v can be reached by traversing exactly three edges from Vertex u, and add an edge connecting Vertices u and v. It is not allowed to add an edge if there is already an edge connecting Vertices u and v.

Find the maximum possible number of edges that can be added.

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i,B_i \leq N

The graph has no self-loops or multiple edges.

The graph is connected.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Find the maximum possible number of edges that can be added.

Sample Input 1

6 5
1 2
2 3
3 4
4 5
5 6

Sample Output 1

4

If we add edges as shown below, four edges can be added, and no more.

Sample Input 2

5 5
1 2
2 3
3 1
5 4
5 1

Sample Output 2

5

Five edges can be added, for example, as follows:

Add an edge connecting Vertex 5 and Vertex 3.

Add an edge connecting Vertex 5 and Vertex 2.

Add an edge connecting Vertex 4 and Vertex 1.

Add an edge connecting Vertex 4 and Vertex 2.

Add an edge connecting Vertex 4 and Vertex 3.

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

def dfs(e)
	stack = [] # stack
	f = Array.new(e.size){false} # is searched flag
	d = Array.new(e.size)
	# init
	s=0
	f[s]=true
	d[s]=0
	e[s].each{|v|
		stack.push([v,s,1]) #v,d to v
	} 
	while(!stack.empty?)
		curv,curp,curd = stack.pop
		f[curv]=true
		d[curv]=curd
		e[curv].each{|nxtv|
			nxtd = curd+1
			if !f[nxtv]
				stack.push([nxtv,curv,nxtd])
			else
				# false if not even cycle
				return [false,[]] if (nxtd-d[nxtv]) %2 != 0
			end
		}
	end
	return [true,d]
end

N,M=gets.chomp.split(' ').map{|n| n.to_i}
E = Array.new(N){[]}
1.upto(M){
	p=gets.chomp.split(' ').map{|n| n.to_i}
	p[0]-=1
	p[1]-=1
	E[p[0]].push(p[1])
	E[p[1]].push(p[0])
}
is_b,d = dfs(E)
if(! is_b)
	puts N*(N-1)/2-M
	exit(0)
end
even = d.inject(0){|sum,d| sum + (d%2==0 ? 1 : 0)}
odd = N-even
puts even*odd - M