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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

typedef struct List {
	struct List *next;
	int v;
} list;

int main()
{
	int i, N, M, u, w;
	list *adj[100001] = {}, e[200001];
	scanf("%d %d", &N, &M);
	for (i = 0; i < M; i++) {
		scanf("%d %d", &u, &w);
		e[i*2].v = w;
		e[i*2+1].v = u;
		e[i*2].next = adj[u];
		e[i*2+1].next = adj[w];
		adj[u] = &(e[i*2]);
		adj[w] = &(e[i*2+1]);
	}
	
	int count = 1, flag[100001] = {}, q[100001], head, tail;
	list *p;
	flag[1] = 1;
	q[0] = 1;
	for (head = 0, tail = 1; head < tail; head++) {
		u = q[head];
		for (p = adj[u]; p != NULL; p = p->next) {
			w = p->v;
			if (flag[w] == 0) {
				flag[w] = -flag[u];
				if (flag[w] == 1) count++;
				q[tail++] = w;
			} else if (flag[w] == flag[u]) break;
		}
		if (p != NULL) break;
	}
	if (head < tail) printf("%lld\n", (long long)N * (N - 1) / 2 - M);
	else printf("%lld\n", (long long)count * (N - count) - M);
	fflush(stdout);
	return 0;
}