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;
using ll = long long;
using ull = unsigned long long;
using P = pair<int,int>;
#define rep(i,n) for(ll i = 0;i < (ll)n;i++)
#define ALL(x) (x).begin(),(x).end()
#define MOD 1000000007

vector<vector<int>> v(100100);
vector<int> col(100100,-1);
ll g[] = {0,0};
bool dfs(int ov,int pre){
  if(col[ov] != -1 && ov != pre){
    if(col[ov] == col[pre])return false;
    else return true;
  }
  bool is = true;
  col[ov] = 1-col[pre];
  g[col[ov]]++;
  for(auto nv: v[ov])is = is && dfs(nv,ov);
  return is;
}

int main(){
  
  ll n,m;
  cin >> n >> m;
  rep(i,m){
    int a,b;cin >> a >> b;
    a--,b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  col[0] = 0;
  bool is = dfs(0,0);
  cout << (is ? g[1]*g[0]-m : n*(n-1)/2-m) << endl;
  


  return 0;
}