Imagine a city with n junctions and m streets. Junctions are numbered from 1 to n.

In order to increase the traffic flow, mayor of the city has decided to make each street one-way. This means in the street between junctions u and v, the traffic moves only from u to v or only from v to u. 

The problem is to direct the traffic flow of streets in a way that maximizes the number of pairs (u, v) where 1 ≤ u, v ≤ n and it is possible to reach junction v from u by passing the streets in their specified direction. Your task is to find out maximal possible number of such pairs.

Input

The first line of input contains integers n and m, (<image>), denoting the number of junctions and streets of the city.

Each of the following m lines contains two integers u and v, (u ≠ v), denoting endpoints of a street in the city.

Between every two junctions there will be at most one street. It is guaranteed that before mayor decision (when all streets were two-way) it was possible to reach each junction from any other junction.

Output

Print the maximal number of pairs (u, v) such that that it is possible to reach junction v from u after directing the streets.

Examples

Input

5 4
1 2
1 3
1 4
1 5


Output

13


Input

4 5
1 2
2 3
3 4
4 1
1 3


Output

16


Input

2 1
1 2


Output

3


Input

6 7
1 2
2 3
1 3
1 4
4 5
5 6
6 4


Output

27

Note

In the first sample, if the mayor makes first and second streets one-way towards the junction 1 and third and fourth streets in opposite direction, there would be 13 pairs of reachable junctions: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (2, 1), (3, 1), (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2000;
vector<int> adj[maxn], Adj[maxn];
bool mark[maxn], make[maxn];
int h[maxn], s[maxn], dp[maxn], par[maxn], cut[maxn][maxn];
void dfs(int v, int Par = -1) {
  mark[v] = true;
  dp[v] = h[v];
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (!mark[u]) {
      h[u] = h[v] + 1;
      dfs(u, v);
      dp[v] = min(dp[v], dp[u]);
    } else if (u != Par)
      dp[v] = min(dp[v], h[u]);
  }
  if (dp[v] == h[v] && Par != -1) cut[v][Par] = cut[Par][v] = -1;
}
int root(int v) { return (par[v] < 0 ? v : par[v] = root(par[v])); }
void Merge(int v, int u) {
  if ((v = root(v)) == (u = root(u))) return;
  if (par[v] > par[u]) swap(v, u);
  par[v] += par[u];
  par[u] = v;
}
int Dfs(int v) {
  mark[v] = true;
  s[v] = -par[v];
  int ret = 0;
  for (int i = 0; i < Adj[v].size(); i++) {
    int u = Adj[v][i];
    if (!mark[u]) {
      ret += Dfs(u);
      s[v] += s[u];
    }
  }
  return ret + (-par[v]) * s[v];
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
    cut[v][u] = cut[u][v] = 1;
  }
  dfs(0);
  memset(par, -1, sizeof par);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (cut[i][j] == 1) Merge(i, j);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (cut[i][j] == -1) {
        int v = root(i), u = root(j);
        Adj[v].push_back(u);
        Adj[u].push_back(v);
      }
  vector<int> ver;
  for (int i = 0; i < n; i++)
    if (par[i] < 0) ver.push_back(i);
  int ans = 0;
  for (int x = 0; x < ver.size(); x++) {
    int v = ver[x];
    memset(mark, 0, sizeof mark);
    int ret = Dfs(v);
    vector<int> coin;
    for (int i = 0; i < Adj[v].size(); i++) {
      int u = Adj[v][i];
      coin.push_back(s[u]);
    }
    memset(make, false, sizeof make);
    make[0] = true;
    for (int j = 0; j < coin.size(); j++)
      for (int i = n - 1; i >= 0; i--) {
        int k = coin[j];
        if (make[i] && i + k < n) make[i + k] = true;
      }
    for (int i = 0; i < n; i++)
      if (make[i]) ans = max(ans, ret + i * (n + par[v] - i));
  }
  cout << ans << endl;
  return 0;
}
