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 N = 2017, M = N * N;
vector<int> adj[N], t[N];
int from[M], to[M], sz[N], cnt = 0, f_h[N], mol[N], n, m;
bool di[N][N];
bool f_mark[N], l_mark[N];
vector<int> topol;
int w[N];
bitset<N> s;
void f_dfs(int v) {
  f_mark[v] = true;
  for (auto u : adj[v]) {
    if (f_mark[u]) continue;
    di[v][u] = true;
    f_h[u] = f_h[v] + 1;
    f_dfs(u);
  }
}
void l_dfs(int v) {
  l_mark[v] = true;
  for (int i = 1; i <= n; i++) {
    if (di[v][i]) {
      if (l_mark[i]) continue;
      l_dfs(i);
    }
  }
  topol.push_back(v);
}
void l_sfd(int v, int id) {
  l_mark[v] = true;
  mol[v] = id;
  sz[id]++;
  for (int i = 1; i <= n; i++) {
    if (di[i][v]) {
      if (l_mark[i]) continue;
      l_sfd(i, id);
    }
  }
}
pair<int, int> g_dfs(int v, int p) {
  pair<int, int> ret = {0, sz[v]};
  for (auto u : t[v]) {
    if (u == p) continue;
    pair<int, int> e = g_dfs(u, v);
    ret.first += e.first;
    ret.first += sz[v] * e.second;
    ret.second += e.second;
  }
  return ret;
}
int solve(int v) {
  int ret = (n - sz[v]) * sz[v], c = 0;
  for (auto u : t[v]) {
    pair<int, int> e = g_dfs(u, v);
    ret += e.first;
    w[c++] = e.second;
  }
  s.reset();
  s[0] = true;
  for (int i = 0; i < c; i++) {
    s |= (s << w[i]);
  }
  int mx = -1;
  for (int i = 0; i <= n - sz[v]; i++) {
    if (s[i]) mx = max(mx, i * (n - sz[v] - i));
  }
  return ret + mx;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> from[i] >> to[i];
    adj[from[i]].push_back(to[i]);
    adj[to[i]].push_back(from[i]);
  }
  f_dfs(1);
  for (int i = 0; i < m; i++) {
    int u = from[i], v = to[i];
    if (di[u][v] || di[v][u]) continue;
    if (f_h[v] > f_h[u])
      di[v][u] = true;
    else
      di[u][v] = true;
  }
  l_dfs(1);
  memset(l_mark, 0, sizeof l_mark);
  reverse(topol.begin(), topol.end());
  for (auto i : topol) {
    if (l_mark[i]) continue;
    l_sfd(i, cnt++);
  }
  for (int i = 0; i < m; i++) {
    int u = from[i], v = to[i];
    if (mol[u] == mol[v]) continue;
    t[mol[u]].push_back(mol[v]);
    t[mol[v]].push_back(mol[u]);
  }
  int ans = 0, ret = 0;
  for (int i = 0; i < cnt; i++) ans += sz[i] * sz[i];
  for (int i = 0; i < cnt; i++) {
    ret = max(ret, solve(i));
  }
  cout << ans + ret << '\n';
}
