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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

struct dsu {
    vector<int> p;
    vector<int> sz;
    int n;

    dsu(int _n) : n(_n) {
        p.resize(n);
        iota(p.begin(), p.end(), 0);
        sz.assign(n, 1);
    }

    inline int get(int x) {
        if (p[x] == x) {
            return x;
        } else {
            return p[x] = get(p[x]);
        }
    }
    inline bool unite(int x, int y) {
        x = get(x);
        y = get(y);
        if (x == y) {
            return false;
        }
        if (sz[x] < sz[y]) {
            swap(x, y);
        }
        p[x] = y;
        sz[y] += sz[x];
        return true;
    }
    inline bool same(int x, int y) {
        return (get(x) == get(y));
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    dsu uf(2 * n);
    for (int i = 0; i < m; i++) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        uf.unite(x, n + y);
        uf.unite(y, n + x);
    }
    int b = 0;
    for (int i = 0; i < 2 * n; i++) {
        if (uf.get(i) == i) {
            b++;
        }
    }
    if (b == 1) {
        ll ans = 1LL * n * (n - 1) / 2;
        ans -= m;
        cout << ans << '\n';
    } else if (b == 2) {
        vector<ll> t(2 * n);
        for (int i = 0; i < n; i++) {
            t[uf.get(i)]++;
        }
        sort(t.rbegin(), t.rend());
        ll ans = t[0] * t[1] - m;
        cout << ans << '\n';
    } else {
        assert(false);
    }
    return 0;
}