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, low[2005], num[2005], cnt, p[2005], br[2005][2005], sz[2005], ans,
    child[2005], res, f[2005][2005];
vector<int> a[2005], b[2005];
vector<pair<int, int> > bridge, edge;
void predfs(int u, int p) {
  low[u] = num[u] = ++cnt;
  for (int i = 0; i < a[u].size(); ++i) {
    int v = a[u][i];
    if (v == p) continue;
    if (num[v])
      low[u] = min(low[u], num[v]);
    else {
      predfs(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] == num[v])
        bridge.push_back(make_pair(u, v)), br[u][v] = br[v][u] = true;
    }
  }
}
int find(int u) { return (u == p[u]) ? u : p[u] = find(p[u]); }
void join(int u, int v) {
  int pu = find(u), pv = find(v);
  if (pu != pv) p[pu] = pv, sz[pv] += sz[pu];
}
void dfs(int u, int p) {
  int v;
  child[u] = sz[u];
  for (int i = 0; i < b[u].size(); ++i)
    if ((v = b[u][i]) != p) {
      dfs(v, u);
      child[u] += child[v];
      res += child[v] * sz[u];
    }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    edge.push_back(make_pair(u, v));
    a[u].push_back(v);
    a[v].push_back(u);
  }
  predfs(1, 1);
  for (int i = 1; i <= n; ++i) p[i] = i, sz[i] = 1;
  for (int i = 0; i < edge.size(); ++i)
    if (!br[edge[i].first][edge[i].second]) join(edge[i].first, edge[i].second);
  for (int i = 0; i < bridge.size(); ++i) {
    int u = find(bridge[i].first), v = find(bridge[i].second);
    b[u].push_back(v);
    b[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i)
    if (p[i] == i) {
      res = 0;
      dfs(i, i);
      for (int j = 1; j <= b[j].size(); ++j)
        for (int k = 0; k <= 2000; ++k) f[j][k] = 0;
      f[0][0] = 1;
      for (int j = 1; j <= b[i].size(); ++j)
        for (int k = 0; k <= 2000; ++k) {
          f[j][k] = f[j - 1][k];
          if (k >= child[b[i][j - 1]])
            f[j][k] |= f[j - 1][k - child[b[i][j - 1]]];
        }
      for (int j = 0; j <= 2000; ++j)
        if (f[b[i].size()][j]) ans = max(ans, res + j * (child[i] - j - sz[i]));
    }
  for (int i = 1; i <= n; ++i)
    if (find(i) == i) ans += sz[i] * (sz[i] - 1);
  cout << ans + n;
}
