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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;

typedef long long int Long;
typedef long double Double;
const int MAXN = 1e5 + 4;

char vis[MAXN];
int n, m, cnt3, ans;
unordered_set<int> vS, A[4][MAXN];
set<pair<int,int>> S[4];
queue<pair<int,int>> Q;

void update(int u, int v) {
	for(int w : A[2][u]) {
		if(w != u && w != v) {
			S[3].emplace(w, v);
		}
	}
	for(int w : A[2][v]) {
		if(w != u && w != v) {
			S[3].emplace(w, u);
		}
	}
}

int main() {
	scanf("%d%d", &n, &m);
	for(int i = 0, u, v; i < m; ++i) {
		scanf("%d%d", &u, &v);
		A[1][u].insert(v);
		A[1][v].insert(u);
	}

	for(int i = 1; i <= n; ++i) {
		for(int v : A[1][i]) {
			for(int w : A[1][v]) {
				if(w != i)
					A[2][i].insert(w);
			}
		}
	}

	for(int i = 1; i <= n; ++i) {
		for(int v : A[1][i]) {
			for(int w : A[2][v]) {
				if(w != i) {
					A[3][i].insert(w);
					S[3].emplace(i, w);
				}
			}
		}
	}

	while(S[3].size()) {
		auto p = *S[3].begin();
		int u = p.first,
			v = p.second;

		S[3].erase(S[3].begin());
		S[3].erase({ v, u });

		if(!A[1][u].count(v)) {
			++ans;
			A[1][u].insert(v);
			A[1][v].insert(u);

			update(u, v);
		}
	}

	printf("%d\n", ans);
	return 0;
}