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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int he[N],ed[N],e,ord[N],Cont[2],n,m;
bool p,P[N];
struct syzs{
	int ne,va,to;
} a[N<<1];
void add(int u,int v){
	e++;
	a[e].to=v;
	a[e].ne=he[u];
	he[u]=e;
}
void dfs(int k,int fa){
	if(p==true) return;
	ed[k]=(ed[fa]+1)%2;
	for(int i=he[k];i;i=a[i].ne){
		int v=a[i].to;
		if(v==fa) continue;
		if(!P[v]){
			P[v]=true;
			Cont[ed[k]]++;
			ord[v]=ed[k];
			dfs(v,k);
		}
		else if(ed[k]!=ord[v]) p=true;
	}
}
int main(){
	cin>>n>>m;
	for(int i=1,u,v;i<=m;i++){
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	ed[0]=P[1]=Cont[1]=1;
	ord[1]=ed[0];
	dfs(1,0);
	if(p==true){
		cout<<n*(n-1)/2-m;
	}
	else cout<<(Cont[0]*Cont[1])-m;
	puts("");
}
/*
8 9
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 5
5 2
*/