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: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int N=100010;
int n,m;
int to[N<<1],edge_tot;
vector<int> point[N];
void add_edge(int f,int t){
	to[edge_tot]=t;
	point[f].push_back(edge_tot++);
	return;
}
int color[N],tot[3];
bool vis[N];
bool fill(int k,int c){
	vis[k]=1,color[k]=c,tot[c]++;
	int x;
	for(int i=0;i<point[k].size();i++){
		x=to[point[k][i]];
		if(!vis[x]){
			if(!fill(x,!c)) return 0;
		}else if(color[x]==color[k]) return 0;
	}
	return 1;
}
int main(){
	int t1,t2;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d%d",&t1,&t2);
		add_edge(t1,t2);add_edge(t2,t1);
	}
	if(fill(1,1)) cout<<1LL*tot[0]*tot[1]-m;
	else cout<<1LL*n*(n-1)/2-m;
	return 0;
}