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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using Edge = pair<int,int>;
using Graph = vector<vector<int> >;
#define INF 1000000007
typedef long long ll;

Graph G;

vector<int> visited;
bool dfs(int v, int p,int c){
    visited[v] = c;
    for(auto e : G[v]){
        if(e == p)continue;
        if(visited[e] == visited[v])return false;
        else if(visited[e] == INF)dfs(e,v,1-c);
    }
    return true;
}

int main(){
    int N,M;cin >> N >> M;
    G.resize(N);
    visited.assign(N,INF);
    ll res = 0;
    ll w = 0,b = 0;
    for(int i = 0 ; i < M ; i++){
        int u,v;cin>>u>>v;
        u--;v--;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    if(dfs(0,-1,0)){
        for(int i = 0 ; i < N ; i++){
            if(visited[i] == 0)w++;
            else b++;
        }
        res = 1LL*b*w-M;
        cout << res << endl;
    }else{
        res = 1LL*N*(N-1)/2-M;
        cout << res << endl;//完全グラフ
    }
    return 0;
}