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: Wrong Answer)

import java.util.*;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		ArrayList<Integer>[] a = new ArrayList[n];
		for(int i=0; i<n; i++){
			a[i] = new ArrayList<Integer>();	
		}
		for(int i=0; i<m; i++){
			int ta = sc.nextInt() - 1;
			int tb = sc.nextInt() - 1;
			a[ta].add(tb);
			a[tb].add(ta);
		}
		sc.close();
		
		int nume = 0, numo = 0;
		boolean[] use = new boolean[n];
		boolean[] odd = new boolean[n];
		boolean[] even = new boolean[n];
		ArrayDeque<Integer[]> que = new ArrayDeque<Integer[]>();
		que.add(new Integer[]{0, 0});
		while(!que.isEmpty()){
			Integer[] cur = que.poll();
			use[cur[0]] = true;
			if(cur[1] == 0){
				numo++;
				odd[cur[0]] = true;
			}else{
				nume++;
				even[cur[0]] = true;
			}
			for(int i=0; i<a[cur[0]].size(); i++){
				Integer next = a[cur[0]].get(i);
				if(cur[1] == 0 && !even[next]){
					que.add(new Integer[]{next, 1});
				}else if(cur[1] == 1 && !odd[next]){
					que.add(new Integer[]{next, 0});
				}
			}
		}

		if(odd[0] && even[0]){
			System.out.println((long)(n-1)*(long)n/2 - m);
		}else{
			long ans = 0;			
			for(int i=0; i<n; i++){
				if(odd[i]){
					ans += numo - a[i].size();
				}else{
					ans += nume - a[i].size();
				}
			}
			System.out.println(ans / 2);
		}
	}
}