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 Python (3.4.3) (judge verdict: Wrong Answer)

from collections import defaultdict
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)

n, m = map(int, (input().split()))
dic = defaultdict(list)
seen = [-1] * (n+1)

for _ in range(m):
    a, b = map(int, input().split())
    dic[a].append(b)
    dic[b].append(a)


def dfs(v, todo, color):
    if seen[v] == color:  # 同じ色で戻ってくる = 偶サイクル
        return
    elif seen[v] == 1-color:  # 異なる色で戻ってくる = 奇サイクル
        return False

    seen[v] = color
    for new_v in todo:
        new_todo = []
        new_todo += dic[new_v]
        new_todo.remove(v)
        new_color = 1 - color
        return dfs(new_v, new_todo, new_color)


if dfs(1, dic[1], 0) is False:
    ans = 0.5 * n * (n-1) - m

else:
    black = seen.count(0)
    white = seen.count(1)
    ans = black * white - m

print(int(ans))