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=1e6;
int n,m;

vector<vector<int>>v;
int vis[N],cyc=0;

void dfs(int u ,int p,int dep){
	if(dep%2)vis[u]=1;
	else vis[u]=0;
	for(int i=0;i<v[u].size();i++){
		int x=v[u][i];
		if(x==p)continue;
		dfs(x,u,dep+1);
	}
	
}
int main() {
	scanf("%d%d",&n,&m);
	v.resize(n+1);
	for(int a,b, i=0;i<m;i++){
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	if(m>n-1){
		long long ans=n* 1LL *(n-1);
		ans/=2;
		ans-=m;
		cout<<ans<<endl;
	}
	else{
		dfs(1,-1,0);
		long long ct1=0,ct2=0;
		for(int i=1;i<=n;i++){
			if(vis[i]==1)ct1++;else
			ct2++;
		}

		long long ans=1LL* ct1*ct2;
		ans-=m;
		cout<<ans<<endl;
	}
	return 0;
}