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 Java7 (OpenJDK 1.7.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ

		Scanner sc = new Scanner(System.in);
		int c = 0;
		int N = sc.nextInt();
		int M = sc.nextInt();
		int A[] = new int[M];
		int B[] = new int[M];
		for (int i = 0; i < M; i++) {
			A[i] = sc.nextInt() - 1;
			B[i] = sc.nextInt() - 1;
		}

		byte br[][] = new byte[N][N];
		for (int i = 0; i < M; i++) {
			br[A[i]][B[i]] = 1;
			br[B[i]][A[i]] = 1;
		}
		byte br2[][] = new byte[N][N];
		byte br3[][] = new byte[N][N];
		int d = -1;

		while (c > d) {
			d = c;
			for (int u = 0; u < N; u++) {
				for (int v = 0; v < N; v++) {
					for (int i = 0; i < N; i++) {
						br2[u][v] += br[u][i] * br[i][v];
					}
				}
			}
			for (int u = 0; u < N; u++) {
				for (int v = 0; v < N; v++) {
					for (int i = 0; i < N; i++) {
						br3[u][v] += br2[u][i] * br[i][v];
					}
				}
			}
			for (int u = 0; u < N; u++) {
				for (int v = 0; v < N; v++) {
					if (u != v && br[u][v] == 0 && br3[u][v] != 0) {
						br[u][v] = 1;
						br[v][u] = 1;
						c++;
					}
				}
			}
		}
		System.out.println(c);
	}
}