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 getchar getchar_unlocked

#define INF 1000000000

inline int nextInt(void) {
	int ch;
	int minus, x;

	do {
		ch = getchar();
	} while (ch < '-');

	if (ch == '-') {
		minus = 1;
		ch = getchar();
	} else
		minus = 0;

	x = 0;
	while (ch >= '0' && ch <= '9') {
		x = x * 10 + (ch & 15);
		ch = getchar();
	}

	return minus ? -x : x;
}

typedef struct Edge {
	int to;
	struct Edge *next;
} Edge;

Edge e[100000 * 2], *adj[100000];
int e_size;
int d[100000];
int cnt[100000];
int queue[100000];
int front, rear;

void add(int from, int to) {
	e[e_size] = (Edge){to, adj[from]};
	adj[from] = e + e_size++;
}

int main(void) {
	int n, m;

	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; i++) {
		const int a = nextInt() - 1, b = nextInt() - 1;
		add(a, b);
		add(b, a);
	}
	for (int 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 (Edge *p = adj[from]; p; p = p->next) {
			const int to = p->to;
			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 (int i = 0; i < 100000; i++)
		sum[i & 1] += cnt[i];

	printf("%lld\n", sum[0] * sum[1] - m);

	return 0;
}