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 <cstdio>
#include <vector>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;

int n,m,u,v;
vector<int> path[maxn];
int c[maxn];
bool f;

void dfs(int p)
{
	for(int i=0;i<path[p].size();i++)
	{
		int tmp=path[p][i];
		if (c[tmp]==-1)
		{
			c[tmp]=1-c[p];
			dfs(tmp);
			if (!f) return;
		}
		else if (c[tmp]==c[p]) 
		{
			f=false;
			return;
		}
	}
	return;
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		path[u].push_back(v);
		path[v].push_back(u);
	}
	
	f=true;
	for(int i=1;i<=n;i++) c[i]=-1;
	c[1]=0;
	dfs(1);
	
	if (!f)
	{
		printf("%lld\n",(LL)n*(n-1)/2-m);
		return 0;
	}
	
	int cnt=0;
	for(int i=1;i<=n;i++) if (c[i]==0) cnt++;
	printf("%lld\n",(LL)cnt*(n-cnt)-m);
	return 0;
}