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: Runtime Error)

#include<bits/stdc++.h>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
const int INF = 1e8;
using namespace std;

const int MAX_V = 100005;


vector<int> g[MAX_V];
int dis[MAX_V];

void bfs(int s, int n){
	queue<int> q;

	rep(i,n) dis[i] = INF;
	dis[s] = 0;
	q.push(s);

	int u;
	while(not q.empty()){
		u = q.front(); q.pop();
		rep(i,g[u].size()){
			int e = g[u][i];
			if(dis[e] == INF){
				dis[e] = dis[u] + 1;
				q.push(e);
			}
		}
	}
}

int solve(int n){
	int maxi = 0;
	int tgt = 0; //?????????s?????????????????????

	bfs(0, n);
	rep(i,n){
		if(dis[i] == INF) continue;
		if(maxi < dis[i]){
			maxi  = dis[i];
			tgt = i;
		}
	}

	bfs(tgt, n);
	maxi = 0; //tgt????????????????????\???????????¢maxi
	int p;
	rep(i,n){
		if(dis[i] == INF) continue;
		if(maxi < dis[i]){
			p = i;
				maxi = max(maxi, dis[i]);
		}
	}

	return maxi;
}


int main(){
	long long n, m;
	cin >> n >> m;


	rep(i,m){
		int a, b;
		cin >> a >> b;
		a--; b--;
		g[a].emplace_back(b);
		g[b].emplace_back(a);
	}

	if(n - m == 1){
		//int p = solve(n);
		//show(p)
		if(n % 2 == 0){
			cout << n / 2 + n - m << endl;
		}else{
			cout << n / 2 + 1 + n - m << endl;
		}
	}else{
		assert(false);

		cout << n * (n - 1) / 2 - m << endl;
	}
}