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;
#define REP(i,a) for(int i = 0; i < (a); i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const long long LINF = 1e18;
const long long MOD = 1e9 + 7;

#define MAX_N 100000

int n,m;
vector<int> G[MAX_N];
int color[MAX_N];

bool dfs(int v, int c){
    color[v] = c;
    for(auto u : G[v]){
        if(color[u] == c) return false;
        if(color[u] == 0 && !dfs(u, -c)){
            return false;
        }
    }
    return true;
}

signed main(){
    cin >> n >> m;
    REP(i,m){
        int a,b;
        cin >> a >> b;
        a--;
        b--;
        G[a].emplace_back(b);
        G[b].emplace_back(a);
    }
    if(!dfs(0, 1)){
        cout << n * (n - 1) / 2 - m << endl;
    }else{
        int black = 0, white = 0;
        for(int v = 0; v < n; v++){
            if(color[v] == 1) black++;
            else white++;
        }
        cout << black * white - m << endl;
    }
    return 0;
}