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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int N=100001;
bool isBipartite = true;
bool visit[N];
vector<vector<int> > edge;
int color[N];map<int, int> dist;
queue<int> q; void bfs(int s){q.push(s);                                  
 dist[s] = 0;   
 color[s]=0;      
while (!q.empty()) {
    int u = q.front(); q.pop();                        // queue: layer by layer!
    for (int j = 0; j < (int)edge[u].size(); j++) {
      int v = edge[u][j];                           // for each neighbors of u
      if (!dist.count(v)) {    // dist.find(v.first) == dist.end() also ok
        dist[v] = dist[u] + 1;
		color[v]=1-color[u]  ;                // v unvisited + reachable
        q.push(v);                              // enqueue v for next step
      }
      else if ((dist[v] % 2) == (dist[u] % 2))              // same parity
        isBipartite = false;
  } }
}
int main(){
	int n,m;
	cin>>n>>m;
	edge.assign(n+1,vector<int>());
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	bfs(1);
	if(isBipartite==0){
		cout<<n*(n-1)/2-m;
	}
	else{
		int cnt1=0,cnt2=0;
		for(int i=1;i<=n;i++){
			if(color[i]){
				cnt1++;
			}
			else{
				cnt2++;
			}
		}
		cout<<cnt1*cnt2-m;
	}
}