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)

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

#define fi first
#define se second
#define pb push_back
#define lson o<<1
#define rson o<<1|1
#define CLR(A, X) memset(A, X, sizeof(A))
#define bitcount(X) __builtin_popcountll(X)
typedef long long LL;
typedef pair<int, int> PII;
const double eps = 1e-10;
const LL MOD = 1e9+7;
const auto INF = 0x3f3f3f3f;
int dcmp(double x) { if(fabs(x) < eps) return 0; return x<0?-1:1; }

const int MAXN = 1e5+5;

vector<int> G[MAXN];
int col[MAXN], cnt = 0;
bool flag = 0;

void dfs(int u) {
    cnt += col[u];
    for(int v:G[u]) {
        if(col[v] == -1) {
            col[v] = col[u]^1;
            dfs(v);
        }
        else if(col[v] == col[u]) flag = 1;
    }
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 0; i < m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        G[u].pb(v); G[v].pb(u);
    }
    CLR(col, -1);
    col[1] = 0;
    dfs(1);
    LL ans;
    if(flag) ans = 1LL*n*(n-1)/2;
    else ans = 1LL*cnt*(n-cnt);
    printf("%lld\n", ans-m);
    return 0;
}