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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <numeric>
#include <string.h>
using namespace std;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = 1<<29;
const long long  INFL = 1e18;
const int MOD = 1000000007;
const int MAX = 210000;
const int NM = 100005;
int N, M, c[3], vis[NM], odd;
vector<int> e[NM]; 

void f(int x, int y) {
    if (vis[x]) {
        odd |= (vis[x] != y);
        return;
    }
    vis[x] = y;
    c[y]++;
    for (auto &i : e[x]) {
        f(i, 3 - y);
    }
    return;
}
void solve() {
    cin >> N >> M;
    for (int i = 0, x, y; i < M; i++) {
        cin >> x >> y;
        e[x].push_back(y);
        e[y].push_back(x);
    }
    odd = 0;
    f(1, 1);
    if (odd) {
        cout << 1LL * N * (N - 1) / 2 - M << endl;
    } else {
        cout << 1LL * c[1] * c[2] - M << endl;
    }
    return;
}
//逆元を求める関数
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++) {
      fac[i] = fac[i-1] * i % MOD;
      inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
      finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
//組み合わせ
long long COM (int n, int k) {
    if(n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n-k] % MOD) % MOD;
}
int main(int args, char* argv[]) {
    solve();
}