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 <algorithm>
#include <iostream>
#include <vector>

using namespace std;

vector<int> neighbor[100'000];
bool isVisited[100'000];
int color[100'000];

bool dfs(int node, int col)
{
    isVisited[node] = true;
    color[node] = col;

    bool result = true;

    for (const int c : neighbor[node]) {
        if (isVisited[c]) {
            result = result && color[node] != color[c];
        } else {
            result = result && dfs(c, 1 - col);
        }
    }

    return result;
}

int main()
{
    int64_t n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        neighbor[a].push_back(b);
        neighbor[b].push_back(a);
    }

    fill(isVisited, isVisited + n, false);
    if (dfs(0, 0)) {
        int64_t cnt = 0;
        for (int i = 0; i < n; i++) {
            cnt += color[i] == 0 ? 1 : 0;
        }
        int64_t result = cnt * (n - cnt) - m;
        cout << result << endl;
    } else {
        cout << n * (n - 1) / 2 - m << endl;
    }

    return 0;
}