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;
int dfs_time;
int p[2010], sz[2010], sub[2010];
int dfn[2010], low[2010];
vector<int> graph[2010];
vector<int> S;
vector<pair<int, int> > bridge;
void tarjan(int t, int fa) {
  S.push_back(t);
  dfn[t] = low[t] = dfs_time++;
  for (auto x : graph[t])
    if (x != fa) {
      if (dfn[x] == -1) {
        tarjan(x, t);
        low[t] = min(low[t], low[x]);
        if (dfn[x] == low[x]) {
          bridge.push_back(make_pair(t, x));
          int k;
          do {
            k = S.back();
            p[k] = x;
            S.pop_back();
          } while (k != x);
        }
      } else
        low[t] = min(low[t], dfn[x]);
    }
}
int find(int x) { return p[x] == x ? x : p[x] = find(p[x]); }
int cnt;
void dfs(int t, int fa, int pre) {
  cnt += sz[t] * pre;
  sub[t] = sz[t];
  for (auto x : graph[t])
    if (x != fa) {
      dfs(x, t, pre + sz[t]);
      sub[t] += sub[x];
    }
}
int dp[2010];
int main() {
  int n, m, u, v;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    u--;
    v--;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs_time = 0;
  memset(dfn, -1, sizeof(dfn));
  tarjan(0, -1);
  while (S.size() > 0) {
    p[S.back()] = 0;
    S.pop_back();
  }
  for (int i = 0; i < n; i++) graph[i].clear(), sz[p[i] = find(i)]++;
  for (auto x : bridge) {
    graph[p[x.first]].push_back(p[x.second]);
    graph[p[x.second]].push_back(p[x.first]);
  }
  int res = 0;
  for (int i = 0; i < n; i++)
    if (p[i] == i) {
      cnt = 0;
      dfs(i, -1, 0);
      for (int k = 0; k <= (n - sz[i]) / 2; k++) dp[k] = 0;
      dp[0] = 1;
      for (auto x : graph[i]) {
        for (int k = (n - sz[i]) / 2 - sub[x]; k >= 0; k--)
          if (dp[k]) dp[k + sub[x]] = 1;
      }
      for (int k = (n - sz[i]) / 2; k >= 0; k--)
        if (dp[k]) {
          cnt += k * (n - sz[i] - k);
          break;
        }
      res = max(res, cnt);
    }
  for (int i = 0; i < n; i++)
    if (p[i] == i) res += sz[i] * sz[i];
  cout << res << endl;
  return 0;
}
