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<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int maxm=1e5+10;
int to[maxm<<1],beg[maxm<<1],nex[maxm<<1],col[maxn][2];
int e=0;
inline int read(){
	int x=0,flag=1;
	char ch=getchar();
	while(!isdigit(ch) && ch!='-')ch=getchar();
	if(ch=='-')flag=-1,ch=getchar();
	while(isdigit(ch))x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return x*flag;
}
inline void add(int x,int y){
	to[++e]=y;
	nex[e]=beg[x];
	beg[x]=e;
}
void dfs(int x,bool p){
	int i;
	col[x][p]=1;
	for(i=beg[x];i;i=nex[i])
		if(!col[to[i]][p^1])
			dfs(to[i],p^1);
}
int main(){
	int i,j,k,m,n;
	n=read();m=read();
	for(i=1;i<=m;i++){
		int x,y;
		x=read();y=read();
		add(x,y);add(y,x);
	}
	dfs(1,0);
	int cnt0,cnt1,cnt2;
	cnt0=cnt1=cnt2=0;
	for(i=1;i<=n;i++){
		if(col[i][0] && col[i][1])cnt0++;
		else if(col[i][0])cnt1++;
		else if(col[i][1])cnt2++;
	}
	ll ans=1ll*(1ll*cnt1*cnt2+1ll*cnt0*cnt2+1ll*cnt0*cnt1+1ll*cnt0*(cnt0-1)/2-m);
	printf("%lld\n",ans);
	return 0;
}