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;
typedef pair<int,int> pii;
int main(){
  int n,m;cin>>n>>m;
  int d[n+1];
  vector<int> v[n+1];
  for(int i=0;i<m;i++){
    int a,b;cin>>a>>b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  fill(d,d+n+1,-1);
  d[1]=2;
  queue<pii> q;
  q.push(make_pair(1,2));
  while(q.size()>0){
    pii p=q.front();q.pop();
    int pf=p.first,ps=p.second;
    //cout<<pf<<endl;
    int y;
    if(ps==1)y=2;
    else y=1;
    for(int i=0;i<v[pf].size();i++){
      int now=v[pf][i];
      if(now==1)continue;
      else if(d[now]==-1){
        d[now]=y;
        q.push(make_pair(now,y));
        continue;
      }
      else if(d[now]==0)continue;
      else{
        if(ps==0){
          d[now]=0;
          q.push(make_pair(now,0));
        }
        else{
          if(ps==d[now]){
            d[now]=0;
            q.push(make_pair(now,0));
          }
          else continue;
        }
      }
    }
  }
  int cnt[3];fill(cnt,cnt+3,0);
  for(int i=1;i<=n;i++){
    cnt[d[i]]++;
    //cout<<d[i]<<endl;
  }
  long long ans=0;
  ans+=(long long)(cnt[0]-1)*cnt[0]/2+(long long)cnt[0]*(cnt[1]+cnt[2])+(long long)cnt[1]*cnt[2];
  if(cnt[0]>0){
    ans=(long long)(n-1)*(n-2)/2+(long long)(n-1);
  }
  ans-=(long long)m;
  cout<<ans<<endl;
}