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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pp pair<pair<ll, ll>,pair<ll, ll>>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define vii vector<int>
#define vll vector<ll>
#define mat vector<vector<ll>>
#define lb lower_bound
#define pb push_back
#define fi first
#define sc second
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep2(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define all(x) x.begin(),x.end()
#define pq priority_queue<ll>
#define pqg priority_queue<ll,vector<ll>,greater<ll>>
#define LB(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define UB(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
#define ERASE(v) sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end())
// #define int ll
using namespace std;
const ll INF = (1 << 30 ) - 1;
const ll LLINF = (1LL << 60LL);
const ll MOD = 1000000007;
const ll mod = 998244353;
const ll MAX = 1100000;
const double pi = acos(-1);
const double eps = 1e-10;
ll dx[4] ={1,0,-1,0} , dy[4] ={0,1,0,-1};
 
template<class T> 
inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> 
inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
struct Timer{
    chrono::system_clock::time_point start, end;
    Timer(){ start = chrono::system_clock::now(); }
    ~Timer(){
        end = chrono::system_clock::now();
        auto msec = chrono::duration_cast<chrono::milliseconds>(end - start).count();
        cerr<<"time : "<<msec<<" ms"<<endl;
    }
};


ll n,m;
vll G[100010];
ll color[100010]={};
bool b[100010];



bool dfs(ll s,ll c){
    color[s]=c;
    for(ll t:G[s]){
        if(color[t]==c) return false;
        if(color[t]==0 && !dfs(t,-c)) return  false;
        }
    return true;
}


bool solve(){
    rep(i,n){
        if(color[i]==0){
            if(!dfs(i,1)){
                return false;
            }
        }
    }
    return true;
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Timer TM;


    cin>>n>>m;
    rep(i,m){
        ll a,b;
        cin>>a>>b;
        a--;b--;
        G[a].pb(b);
        G[b].pb(a);
    }
    if(solve()){
        ll a=0,b=0;
        rep(i,n){
            if(color[i]==1) a++;
            else b++;
        }
        cout<<a*b-m<<endl;
    }
    else{
        cout<<n*(n-1)/2-m<<endl;
    }
    return 0;
}