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 <iostream>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <vector>
#include <complex>
#include <utility>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <bitset>
#include <ctime>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cassert>
#include <cstddef>
#include <iomanip>
#include <numeric>
#include <tuple>
#include <sstream>
#include <fstream>

using namespace std;
#define REP(i, n) for (int (i) = 0; (i) < (n); (i)++)
#define FOR(i, a, b) for (int (i) = (a); (i) < (b); (i)++)
#define RREP(i, a) for(int (i) = (a) - 1; (i) >= 0; (i)--)
#define FORR(i, a, b) for(int (i) = (a) - 1; (i) >= (b); (i)--)
#define DEBUG(C) cerr << #C << " = " << C << endl;
using LL = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<LL>;
using VVL = vector<VL>;
using VD = vector<double>;
using VVD = vector<VD>;
using PII = pair<int, int>;
using PDD = pair<double, double>;
using PLL = pair<LL, LL>;
using VPII = vector<PII>;
template<typename T> using VT = vector<T>;
#define ALL(a) begin((a)), end((a))
#define RALL(a) rbegin((a)), rend((a))
#define SORT(a) sort(ALL((a)))
#define RSORT(a) sort(RALL((a)))
#define REVERSE(a) reverse(ALL((a)))
#define MP make_pair
#define FORE(a, b) for (auto &&a : (b))
#define FIND(s, e) ((s).find(e) != (s).end())
#define EB emplace_back
template<typename T> inline bool chmax(T &a, T b){if (a < b){a = b;return true;}return false;}
template<typename T> inline bool chmin(T &a, T b){if (a > b){a = b;return true;}return false;}

const int INF = 1e9;
const int MOD = INF + 7;
const LL LLINF = 1e18;

const int MAX = 1e5 + 10;
int n, m;
set<int> G[MAX];

LL ans = 0;
set<PII> se;
set<PII> u2;

void dfs(int now, int pre, int pre2, int pre3) {
    //if (min(now, pre3)+1==3) cerr << min(now, pre3)+1 << " " << max(now, pre3)+1 << endl;
    if (pre3 != -1 && G[now].count(pre3) == 0  && now != pre3 && se.count(MP(min(now, pre3), max(now, pre3))) == 0) {
        se.emplace(MP(min(now, pre3), max(now, pre3)));
    }
    FORE(e, G[now]) {
        if (e != pre && u2.count(MP(e, pre3)) == 0) {
            u2.emplace(MP(e, pre3));
            dfs(e, now, pre, pre2);
        }
    }
}

int main(void) {
    cin >> n >> m;
    REP(i, m) {
        int a, b;
        cin >> a >> b;
        a--;b--;
        G[a].emplace(b);
        G[b].emplace(a);
    }
    REP(i, 3) {
        REP(j, n) {
            dfs(j, -1, -1, -1);
        }
        FORE(e, se) {
            //cerr << e.first+1 << " " << e.second+1 << " " << i + 1 << endl;
            G[e.first].emplace(e.second);
            G[e.second].emplace(e.first);
            ans++;
        }
        se.clear();
        u2.clear();
    }
    cout << ans << endl;
}