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>
#define REP(i,n) for (int i=0;i<(n);i++)
#define FOR(i,s,e) for (int i=s;i<(e);i++)
#define ALL(v) (v).begin(),(v).end()
#define MP(a,b) make_pair(a,b)
using namespace std;
typedef long long llint;
typedef pair<int, int> P;
const int MOD = (int)1e9 + 7;
const int INF = 999999999;
int v[100001][3];

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, M;
    cin >> N >> M;
    memset(v, 0, sizeof(v));
    REP(i, M){
        int a,b;
        cin >> a >> b;
        if(b-a == 1){
            v[b][0] = 1;
            v[a][1] = 1;
        }else if(b-a == -1){
            v[a][0] = 1;
            v[b][1] = 1;
        }else if(abs(b-a) == 2){
            int c = (b+a)/2;
            v[c][2] = 1;
        }
    }

    bool flag = false;
    FOR(i, 1, N+1){
        if(v[i][0] && v[i][1] && v[i][2]){
            flag = true;
            break;
        }
    }
    int ans = 0;
    int i = 1;
    while(N-i > 0){
        ans += N-i;
        i += (flag ? 1:2);
    }
    ans -= M;
    cout << ans << "\n";
    return 0;
}