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 long long maxn = 2e3 + 100;
const long long maxm = 2e6;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
long long n, m, t, c[maxn], d[maxn], h[maxn], s[maxn], sz[maxn], ans, x;
bool vis[maxn], val[maxm];
bitset<maxn> b;
vector<pair<long long, long long> > g[maxn];
vector<long long> g2[maxn];
void dfs1(long long v, long long e) {
  long long u, w;
  vis[v] = 1;
  h[v] = inf;
  for (long long i = 0; i < g[v].size(); i++) {
    u = g[v][i].first;
    w = g[v][i].second;
    if (w == e) continue;
    if (vis[u])
      h[v] = min(h[v], d[u]), val[w] = 1;
    else {
      d[u] = d[v] + 1;
      dfs1(u, w);
      if (h[u] <= d[v]) val[w] = 1;
      h[v] = min(h[v], h[u]);
    }
  }
  return;
}
void dfs2(long long v) {
  long long u, w;
  vis[v] = 1;
  c[v] = t;
  s[t]++;
  for (long long i = 0; i < g[v].size(); i++) {
    u = g[v][i].first;
    w = g[v][i].second;
    if (val[w] && !vis[u]) dfs2(u);
  }
  return;
}
void dfs3(long long v) {
  long long u;
  vis[v] = 1;
  x += s[v] * (s[v] + d[v]);
  sz[v] = s[v];
  for (long long i = 0; i < g2[v].size(); i++) {
    u = g2[v][i];
    if (!vis[u]) {
      d[u] = d[v] + s[v];
      dfs3(u);
      sz[v] += sz[u];
    }
  }
  return;
}
int main() {
  long long v, u;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> v >> u;
    g[v].push_back({u, i + 1});
    g[u].push_back({v, i + 1});
  }
  dfs1(1, 0);
  fill(vis, vis + maxn, 0);
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      t++;
      dfs2(i);
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < g[i].size(); j++) {
      v = g[i][j].first;
      u = g[i][j].second;
      if (!val[u]) g2[c[i]].push_back(c[v]);
    }
  }
  for (long long i = 1; i <= t; i++) {
    fill(vis, vis + maxn, 0);
    d[i] = 0;
    x = 0;
    dfs3(i);
    b.reset();
    b[0] = 1;
    for (long long j = 0; j < g2[i].size(); j++) {
      v = g2[i][j];
      b |= (b << sz[v]);
    }
    for (long long j = 0; j < n; j++) {
      if (b[j]) ans = max(ans, x + j * (n - s[i] - j));
    }
  }
  cout << ans;
  return 0;
}
