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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int ta[100010],nt[200010],to[200010],co[200010];
void mkls(int v,int e,int *a,int *b,int *c){
  int i;
  for(i=0;i<v;i++)ta[i+1]=-1;
  for(i=0;i<e;i++){
    nt[i  ]=ta[to[i+e]=a[i]];
    nt[i+e]=ta[to[i  ]=b[i]];
    co[ta[b[i]]=i+e]=co[ta[a[i]]=i]=c[i];
  }
}
int main(){
  int n,m,d[100010],i,r,t,a[100010],b[100010],c[100010],q[100010];
  long long ev=0,od=0,ans;
  scanf("%d %d",&n,&m);
  for(i=0;i<n;i++)d[i+1]=-1;
  for(i=0;i<m;i++){
    scanf("%d %d",&a[i],&b[i]);
    c[i]=1;
  }
  mkls(n,m,a,b,c);
  q[t=0]=1;
  d[1]=0;
  for(r=1;r-t;t++){
    for(i=ta[q[t]];i+1;i=nt[i]){
      if(d[to[i]]==-1){
	d[to[i]]=d[q[t]]+1;
	q[r++]=to[i];
      }
      else if(d[q[t]]%2==d[to[i]])goto END;
    }
  }
 END:;
  if(r==t){
    for(i=0;i<n;i++){
      if(d[i+1]%2)od++;
      else      ev++;
    }
    ans=od*ev;//printf("no\n");
  }
  else ans=n*(n-1LL)/2;
  printf("%lld\n",ans-m);
  return 0;
}