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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
 
#define INIT_CAP 10
#define INF 1000000000
 
typedef struct {
	int *elem, cap, size;
} VecInt;
 
VecInt adj[100000];
int d[100000];
int cnt[100000];
int queue[100000];
int front, rear;
 
void init(int n) {
	while (n--)
		adj[n] = (VecInt){malloc(sizeof(int) * INIT_CAP), INIT_CAP, 0};
}
 
void add(int from, int to) {
	if (adj[from].size == adj[from].cap) {
		adj[from].cap <<= 1;
		adj[from].elem = realloc(adj[from].elem, sizeof(int) * adj[from].cap);
	}
	adj[from].elem[adj[from].size++] = to;
}
 
int main(void) {
	int i;
	int n, m;
 
	scanf("%d %d", &n, &m);
 
	init(n);
	for (i = 0; i < m; i++) {
		int a, b;
 
		scanf("%d %d", &a, &b);
		a--; b--;
 
		add(a, b);
		add(b, a);
	}
	for (i = 0; i < n; i++)
		d[i] = INF;
 
	d[0] = 0;
	queue[rear++] = 0;
	while (front != rear) {
		const int from = queue[front++];
		cnt[d[from]]++;
		for (i = 0; i < adj[from].size; i++) {
			const int to = adj[from].elem[i];
			if (d[to] == INF) {
				d[to] = d[from] + 1;
				queue[rear++] = to;
			} else if (d[to] == d[from]) {
				printf("%lld\n", (long long)n * (n - 1) / 2 - m);
				return 0;
			}
		}
	}
 
	long long sum[2] = {0};
	for (i = 0; i < 100000; i++)
		sum[i & 1] += cnt[i];
 
	printf("%lld\n", sum[0] * sum[1] - m);
 
	return 0;
}