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: Runtime Error)

# 二部グラフ判定
def dfs(u, c):
    color[u] = c  # 訪問したら指定の色で塗る
    for v in G[u]:  # uからつながっている頂点を順に取り出す
        if color[v] != -1:  # すでに訪問していた場合
            if color[v] == c:
                return False  # 隣接する頂点が同じ色のときはFalse
            else:
                continue
        else:  # 未訪問の時は色は違う色にしてDFSで再帰的に探索
            if not dfs(v, 1 - c):  # FalseだったらFalseを返す
                return False
    return True


# 頂点数と辺数を入力
N, M = map(int, input().split())
# グラフの入力
G = [[] for _ in range(N)]  # 頂点数Nだけ準備
color = [-1] * N  # 未訪問は-1とする
for _ in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    G[a].append(b)
    G[b].append(a)
# 探索
flag = True
for i in range(N):
    if color[i] == -1:
        if not dfs(i, 0):
            flag = False
            break
if flag:
    ans = color.count(1) * color.count(0) - M
else:
    ans = N * (N - 1) // 2 - M
print(ans)