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 c = 2002;
int n, m, maxi, pl, si[c], rf[c], ki[c], eler[c], visz[c], f[c], cnt, ans;
vector<int> sz[c], sz2[c];
bool v[c], ks[c];
int holvan(int a) {
  if (!ki[a]) {
    return a;
  }
  int x = holvan(ki[a]);
  ki[a] = x;
  return x;
}
void unio(int a, int b) {
  a = holvan(a), b = holvan(b);
  if (a != b) {
    ki[a] = b;
    si[b] += si[a], si[a] = 0;
  }
}
void dfs(int a) {
  v[a] = true, cnt++;
  eler[a] = cnt, visz[a] = cnt;
  for (int x : sz[a]) {
    if (!v[x]) {
      f[x] = a;
      dfs(x);
      visz[a] = min(visz[a], visz[x]);
      if (visz[x] != eler[x]) {
        unio(a, x);
      }
    } else {
      if (f[a] != x) {
        visz[a] = min(visz[a], eler[x]);
      }
    }
  }
}
void dfs2(int a, bool s) {
  v[a] = true;
  rf[a] += si[a];
  for (int x : sz2[a]) {
    if (!v[x]) {
      dfs2(x, 0);
      ans += si[a] * rf[x];
      rf[a] += rf[x];
      if (s) {
        for (int i = n; i >= 0; i--) {
          if (ks[i]) {
            ks[i + rf[x]] = 1;
          }
        }
      }
    }
  }
  if (s) {
    int add = 0;
    for (int i = 0; i < n; i++) {
      if (ks[i]) {
        add = max(add, i * (n - si[a] - i));
      }
    }
    ans += add;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    si[i] = 1;
  }
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    sz[a].push_back(b), sz[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    pl += si[i] * si[i];
    for (int j : sz[i]) {
      int a = holvan(i), b = holvan(j);
      if (a != b) {
        sz2[a].push_back(b);
      }
    }
  }
  ks[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      v[j] = 0, rf[j] = 0, ks[j] = 0;
    }
    ans = 0;
    dfs2(i, 1);
    maxi = max(maxi, ans);
  }
  cout << maxi + pl << "\n";
  return 0;
}
