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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 2002;
const long long mod = 1e9 + 7;
const long double PI = 4 * atan((long double)1);
long long pw(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * pw(a * a % md, b / 2, md) % md
                      : pw(a * a % md, b / 2, md) % md));
}
int n, res, m;
vector<int> adj[maxn], vec[maxn];
int mn, mx, cnt = 0;
int low[maxn], srt[maxn], ch[maxn];
int par[maxn], sz[maxn], ans;
bool f[maxn][maxn], mark[maxn];
bool ct[maxn][maxn];
vector<pair<int, int> > cct;
pair<int, int> e[(int)3e5 + 1];
int getpar(int v) { return (((v == par[v]) ? v : par[v] = getpar(par[v]))); }
void merge(int u, int v) {
  u = getpar(u), v = getpar(v);
  if (u == v) {
    return;
  }
  par[u] = v;
  sz[v] += sz[u];
}
void dfs(int v, int pr) {
  mark[v] = 2;
  ch[v] = sz[v];
  for (auto u : vec[v]) {
    if (u == pr) {
      continue;
    }
    dfs(u, v);
    ch[v] += ch[u];
    res += ch[u] * sz[v];
  }
}
void init(int v, int pr) {
  low[v] = srt[v] = ++cnt;
  for (int u : adj[v]) {
    if (u == pr) {
      continue;
    }
    if (srt[u]) {
      low[v] = min(low[v], srt[u]);
    } else {
      init(u, v);
      low[v] = min(low[v], low[u]);
      if (low[u] == srt[u]) {
        cct.push_back(make_pair(u, v));
        ;
        ct[u][v] = ct[v][u] = 1;
      }
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    ;
    adj[v].push_back(u);
    ;
    e[i] = {u, v};
  }
  init(1, 1);
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  for (int i = 1; i <= m; i++) {
    if (!ct[e[i].first][e[i].second]) {
      merge(e[i].first, e[i].second);
    }
  }
  for (auto i : cct) {
    int u = getpar(i.first), v = getpar(i.second);
    vec[u].push_back(v);
    ;
    vec[v].push_back(u);
    ;
  }
  for (int i = 1; i <= n; i++) {
    if (par[i] != i) {
      continue;
    }
    res = 0;
    dfs(i, i);
    memset(mark, 0, sizeof mark);
    ;
    for (int j = 1; j <= vec[j].size(); j++) {
      for (int k = 0; k < maxn; k++) {
        f[j][k] = 0;
      }
    }
    f[0][0] = 1;
    for (int j = 1; j <= vec[i].size(); j++) {
      for (int k = 0; k < maxn; k++) {
        f[j][k] = f[j - 1][k];
        if (k >= ch[vec[i][j - 1]]) {
          f[j][k] |= f[j - 1][k - ch[vec[i][j - 1]]];
        }
      }
    }
    for (int j = 0; j < maxn; j++) {
      if (f[vec[i].size()][j]) {
        ans = max(ans, res + j * (ch[i] - j - sz[i]));
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (getpar(i) == i) {
      ans += sz[i] * sz[i] - sz[i];
    }
  }
  mx = ans;
  cout << mx + n;
  return (0);
}
