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.Scanner;
import java.util.ArrayList;
import java.util.Arrays;
public class Main{
	static ArrayList<Integer>[] map;
	static int[] ch;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		map=new ArrayList[n];
		ch=new int[n];
		for(int i=0; i<n; i++){
			map[i]=new ArrayList<>();
		}
		for(int i=0; i<m; i++){
			int a =sc.nextInt()-1;
			int b =sc.nextInt()-1;
			map[a].add(b);
			map[b].add(a);
		}
		Arrays.fill(ch,2);
		if(dfs(0,0)){
			long count=0;
			for(int i=0; i<n; i++){
				if(ch[i]==0){
					count++;
				}
			}
			System.out.println(count*(n-count)-m);
		}else{
			System.out.println((long)n*(n-1)/2-m);
		}
	}
	static boolean dfs(int now,int count){
		if(ch[now]==2){
			ch[now]=count%2;
		}else{
			if(ch[now]!=count%2){
				return false;
			}else{
				return true;
			}
		}
		for(int i:map[now]){
			if(!dfs(i,count+1)){
				return false;
			}
		}
		return true;
	}
}