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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#define MaxN 100000+100
using namespace std;
int point[MaxN][2];
vector<int>Ve[MaxN];
void DFS(int Point,int jo)
{
    if(point[Point][jo]==1)
        return ;
    point[Point][jo]=1;
    for(int i=0;i<Ve[Point].size();i++)
        DFS(Ve[Point][i],1-jo);
}
int main()
{
    int N,M;
    scanf("%d %d",&N,&M);
    for(int i=0;i<M;i++)
    {
        int a,b;
        scanf("%d %d",&a,&b);
        Ve[a].push_back(b);
        Ve[b].push_back(a);
    }
    DFS(1,1);
    int a,b,c;
    a=b=c=0;
    for(int i=1;i<=N;i++)
    {
        if(point[i][0] && point[i][1])
            c++;
        else
            if(point[i][0])
                a++;
            else
                b++;
    }
    printf("%d\n",c*a+c*b+c*(c-1)/2+a*b-M);
    return 0;
}