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 PyPy3 (2.4.0) (judge verdict: Accepted)

from collections import deque
N, M = map(int, input().split())
G = [[] for i in range(N)]
for m in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    G[a].append(b)
    G[b].append(a)

color = [-1 for i in range(N)]
visits = [False for i in range(N)]
paths = deque()


def Bipartial():
    paths.append(0)
    color[0] = 0
    while paths:
        x = paths.pop()
        visits[x] = True
        for g in G[x]:
            if visits[g] and color[g] == color[x]:
                return False
            elif not visits[g]:
                paths.append(g)
                color[g] = (color[x]+1) % 2
    return True


if Bipartial():
    B = color.count(0)
    W = color.count(1)
    ans = B * W - M
else:
    ans = (N * (N - 1)) // 2 - M
print(ans)