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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include <bitset>
#include<math.h>
using namespace std;
#define INF 110000000000
#define MOD 1000000007
typedef long long ll;
typedef pair<int,int> P;

vector<int> V[110000];
int color[110000];

//二部グラフならtrueを返す
bool dfs(int x,int c){
  return true;
  color[x]=c;
  for(int i=0;i<(int)V[x].size();i++){
    //隣接している頂点が同じ色ならfalse
    if(color[V[x][i]]==c) return false;
    //隣接している頂点がまだ塗られていないなら-cで塗る
    if(color[V[x][i]]==0 && !dfs(V[x][i],-c)) return false;
  }
  return true;
}

int main(){
  int N,M,a,b;
  cin>>N>>M;

  for(int i=0;i<M;i++){
    cin>>a>>b;
    V[a].push_back(b);
    V[b].push_back(a);
  }

  if(dfs(1,1)){
    int countA=0,countB=0;
    for(int i=1;i<=N;i++){
      if(color[i]==1) countA++;
      else countB++;
    }
    cout<<countA*countB-M<<endl;
  }else{
    cout<<N*(N-1)/2-M<<endl;
  }
}