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 h[2010], n, m, mn[2010], cnt[2010], dp[2010];
vector<int> g[2010], gc[2010];
bitset<2010> bt, b;
void dfs1(int np, int dep) {
  mn[np] = h[np] = dep;
  bt[np] = 1;
  for (auto &v : g[np]) {
    if (!bt[v]) dfs1(v, dep + 1);
    if (h[v] - h[np] + 1 != 0) mn[np] = min(mn[np], mn[v]);
  }
}
void dfs2(int np, int c) {
  static int col = 1;
  cnt[c]++;
  bt[np] = 1;
  for (auto &v : g[np]) {
    if (!bt[v]) {
      if (mn[v] <= h[np])
        dfs2(v, c);
      else {
        gc[c].push_back(col);
        gc[col].push_back(c);
        dfs2(v, col++);
      }
    }
  }
}
int dfs3(int np, int fath) {
  int ans = dp[np] = 0;
  for (auto &v : gc[np])
    if (v != fath) ans += dfs3(v, np), dp[np] += dp[v];
  dp[np] += cnt[np];
  ans += dp[np] * cnt[np];
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, ti, tj; i <= m; i++) {
    scanf("%d%d", &ti, &tj);
    ti--;
    tj--;
    g[ti].push_back(tj);
    g[tj].push_back(ti);
  }
  int ans = 0;
  dfs1(0, 0);
  bt.reset();
  dfs2(0, 0);
  for (int i = 0; cnt[i]; i++) {
    int res = dfs3(i, -1);
    b.reset();
    b.set(0);
    for (auto &v : gc[i]) b |= b << dp[v];
    for (int j = 0; j < n; j++)
      if (b[j]) ans = max(ans, res + j * (n - cnt[i] - j));
  }
  printf("%d\n", ans);
  return 0;
}
