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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    static List<Integer>[] graph;
    static int[] color;
    static long wCnt = 0;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        graph = new List[N+1];
        for (int i = 1; i <= N; i++) graph[i] = new ArrayList<Integer>();
        for (int i = 0; i < M; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            graph[a].add(b);
            graph[b].add(a);
        }
        long ans = 0;
        color = new int[N+1];
        if (dfs(0, 1, 1)) {
            ans = wCnt * (N - wCnt) - M;
        } else{
           ans = ((long) N * (N - 1)) / 2 - M;
        }
        System.out.println(ans);
        sc.close();

    }

    private static boolean dfs(int pre, int cur, int c) {
        if(color[cur] != 0){
            return color[cur] == c;
        }
        color[cur] = c;
        if(c == 1) wCnt++;
        boolean ret = true;
        for(int i = 0; i < graph[cur].size(); i++){
            int x = graph[cur].get(i);
            if(x == pre) continue;
            ret &= dfs(cur, x, -c);
        }

        return ret;
    }

}