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

#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline int read(){
    int s=0,w=1;
    char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}
    while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();}
    return s*w;
}
const int maxn=1e5+10;
struct edge{
    int to,nxt;
}e[maxn*2];
int head[maxn],tot,col[maxn];
ll c1,c2;
int flag;
void addedge(int u,int v){
    e[++tot].to=v;
    e[tot].nxt=head[u];
    head[u]=tot;
}
void dfs(int u,int color){
    if(flag) return;
    col[u]=color;
    if(color==1) ++c1;
    else ++c2;
    for(int i=head[u];i;i=e[i].nxt)
    {
        int v=e[i].to;
        if(!col[v]) dfs(v,3-color);
        else if(col[v]==color){
            flag=true;
            return;
        }
    }
}
ll n,m;
int main(){
    n=read(),m=read();
    for(int i=1;i<=m;i++){
        int u,v;
        u=read(),v=read();
        addedge(u,v);
        addedge(v,u);
    }
    dfs(1,1);
    if(flag) printf("%lld",1LL*n*(n-1)/2-m);
    else printf("%lld",c1*c2-m);

    return 0;
}