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 n, m, counter, tot;
int low[2013];
int num[2013];
set<int> adj[2013];
vector<pair<int, int> > bridges;
int can[2013];
vector<int> has;
int p[2013];
int s[2013];
int find(int x) { return p[x] == x ? x : p[x] = find(p[x]); }
void merge(int a, int b) {
  if (find(a) == find(b)) return;
  s[find(b)] += s[find(a)];
  p[find(a)] = find(b);
}
void dfs(int x, int pp) {
  num[x] = low[x] = counter++;
  for (int i : adj[x]) {
    if (!num[i]) {
      dfs(i, x);
      low[x] = min(low[x], low[i]);
      if (low[i] > num[x]) bridges.push_back(make_pair(x, i));
    } else if (i != pp)
      low[x] = min(low[x], num[i]);
  }
}
int dfs2(int x, int pp, int cur) {
  tot += s[x];
  int res = cur * s[x];
  for (int i : adj[x])
    if (i != pp) res += dfs2(i, x, cur + s[x]);
  return res;
}
int sc[2013];
int findC(int x = find(1), int pp = 0) {
  int ok = 1;
  sc[x] = s[x];
  for (int i : adj[x])
    if (i != pp) {
      int pos = findC(i, x);
      if (pos) return pos;
      if (sc[i] > n / 2) ok = 0;
      sc[x] += sc[i];
    }
  if (n - sc[x] > n / 2) ok = 0;
  return ok ? x : 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].insert(v);
    adj[v].insert(u);
  }
  counter = 1;
  dfs(1, 0);
  for (pair<int, int> e : bridges) {
    adj[e.first].erase(e.second);
    adj[e.second].erase(e.first);
  }
  for (int i = 1; i <= n; i++) p[i] = i, s[i] = 1;
  for (int i = 1; i <= n; i++)
    for (int j : adj[i]) merge(i, j);
  for (int i = 1; i <= n; i++) adj[i].clear();
  for (pair<int, int> e : bridges) {
    adj[find(e.first)].insert(find(e.second));
    adj[find(e.second)].insert(find(e.first));
  }
  int ans = 0;
  for (int c = findC(); c <= n; c += n)
    if (find(c) == c) {
      int cur = 0;
      for (int i = 1; i <= n; i++)
        if (find(i) == i) cur += s[i] * s[i];
      has.clear();
      for (int i : adj[c]) {
        tot = 0;
        cur += dfs2(i, c, s[c]);
        has.push_back(tot);
      }
      for (int i = 0; i <= n; i++) can[i] = 0;
      can[0] = 1;
      for (int j : has) {
        for (int i = n; i >= 0; i--)
          if (i + j <= n) can[i + j] |= can[i];
      }
      int best = 0;
      for (int i = 1; i <= n; i++)
        if (can[i]) best = max(best, i * (n - s[c] - i));
      cur += best;
      ans = max(ans, cur);
    }
  printf("%d\n", ans);
  return 0;
}
