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 + 10;
const int maxm = 2000 * 1000 + 10;
int n, m;
pair<int, int> e[maxm];
bool cut[maxm];
vector<pair<int, int> > adj[maxn];
bool mark[maxn];
int h[maxn];
int cm, comp[maxn];
vector<int> comp2[maxn];
int ans = 0, ans2 = 0;
bool ekh[maxn];
vector<int> adj2[maxn];
int sz[maxn];
int dfs_cut(int v, int par);
void dfs_comp(int v);
void chk(int v);
int dfs_size(int v);
void add(int u);
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    e[i] = pair<int, int>(u, v);
    adj[u].push_back(pair<int, int>(v, i));
    adj[v].push_back(pair<int, int>(u, i));
  }
  dfs_cut(0, -1);
  fill(mark, mark + n, false);
  for (int i = 0; i < n; i++) {
    if (!mark[i]) {
      dfs_comp(i);
      cm++;
    }
  }
  for (int i = 0; i < m; i++) {
    if (cut[i]) {
      adj2[comp[e[i].first]].push_back(comp[e[i].second]);
      adj2[comp[e[i].second]].push_back(comp[e[i].first]);
    }
  }
  for (int i = 0; i < cm; i++) chk(i);
  for (int i = 0; i < cm; i++)
    ans += (comp2[i].size() * ((int)comp2[i].size()));
  cout << ans << endl;
  return 0;
}
int dfs_cut(int v, int par) {
  mark[v] = true;
  int res = h[v];
  for (auto u2 : adj[v]) {
    int u = u2.first;
    if (!mark[u]) {
      h[u] = h[v] + 1;
      int tmp = dfs_cut(u, v);
      if (tmp >= h[u]) {
        cut[u2.second] = true;
      }
      res = min(res, tmp);
    } else {
      if (u != par) res = min(res, h[u]);
    }
  }
  return res;
}
void dfs_comp(int v) {
  mark[v] = true;
  comp[v] = cm;
  comp2[cm].push_back(v);
  for (auto u2 : adj[v]) {
    int u = u2.first;
    int c = u2.second;
    if (cut[c] || mark[u]) continue;
    dfs_comp(u);
  }
}
void chk(int v) {
  fill(mark, mark + cm, false);
  ans2 = 0;
  dfs_size(v);
  fill(ekh, ekh + n, false);
  ekh[0] = true;
  for (int u : adj2[v]) add(u);
  for (int i = 0; i < n; i++)
    if (ekh[i]) {
      ans = max(ans, ans2 + (n - i - (int)comp2[v].size()) *
                                (n + i - (int)comp2[v].size()) / 4);
      return;
    }
}
int dfs_size(int v) {
  sz[v] = comp2[v].size();
  mark[v] = true;
  int res = 0;
  for (auto u : adj2[v]) {
    if (!mark[u]) res += dfs_size(u);
  }
  ans2 += sz[v] * res;
  sz[v] += res;
  return sz[v];
}
void add(int u) {
  bool ekh2[maxn] = {};
  for (int i = 0; i <= n; i++)
    if (ekh[i]) ekh2[abs(i - sz[u])] = true, ekh2[abs(i + sz[u])] = true;
  for (int i = 0; i <= n; i++) ekh[i] = ekh2[i];
}
