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<cstdio>
#include<cstring>
#include <cstdlib>  
#include <math.h>
#include <cmath>
#include<complex>
#include<cctype>
#include<string>
#include<set>
#include<iomanip>
#include <map>
#include<algorithm>
#include <functional>
#include<vector>
#include<climits>
#include<stack>
#include<queue>
#include<bitset>
#include <deque>
#include <climits>
#include <typeinfo>
#include <utility> 
using namespace std;
using Complex = complex<double>;
using ll = long long;
template<typename T>using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const ll inf = 1145141919810364364LL;
double PI = acos(-1.0);
#define all(x) (x).begin(),(x).end()
#define puts(x) cout << x << endl
#define rep(i,m,n) for(ll i = m;i < n;++i)
#define pb push_back
#define fore(i,a) for(auto &i:a)
#define rrep(i,m,n) for(ll i = m;i >= n;--i)

ll N, M;
int col[101010];
vector<int>E[101010];
int fl = 0;
void dfs(int now) {
	fore(x, E[now]) {
		if (col[x] == -1) {
			col[x] = 1-col[now];
			dfs(x);
		}
		else if (col[x] == col[now])fl = 1;
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		E[a].pb(b);
		E[b].pb(a);
	}
	for (int i = 0; i < 101010; i++)col[i] = -1;
	col[0] = 0;
	dfs(0);

	ll w = 0, b = 0;
	for (int i = 0; i < N; i++)((col[i] == 0) ? w : b) += 1;
	if (fl)puts(N*(N - 1LL) / 2LL - M);
	else puts(N*(N-1LL)/2LL - M - w*(w-1)/2LL - b*(b-1)/2LL);
	
	return 0;
}