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 Graph = vector<vector<int>>;
Graph G;
vector<int>color;

bool dfs(int v, int c)
{
    color[v] = c;
    for(auto nx: G[v])
    {
        if(color[nx] != -1)
        {
            if(color[nx] == c)
            {
                return false;
            }
            continue;
        }
        if(!dfs(nx,1-c))
        {
            return false;
        }
    }
    return true;
}
int main()
{
    int n,m;
    cin >> n >> m;
    G.resize(n);
    for(int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        a--,b--;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    color.assign(n,-1);
    long long int ans;
    if(dfs(0,0))
    {
        int c = 0;
        int d = 0;
        for(auto x: color)
        {
            if(x == 0)
            {
                c++;
            }
            else
            {
                d++;
            }
        }
        ans = c*d - m;
        cout << ans << endl;
    }
    else
    {
        long long int ans1 = n;
        long long int ans2 = n - 1;
        long long int ans3 = ans1 * ans2;
        cout << ans3 / 2 - m << endl;
    }
    return 0;
}