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, indexx[2005], low[2005], belong[2005];
int cnt, bcc;
int sz[2005], s[2005], dp[2005], f[2005][2005];
stack<int> st;
vector<int> a[2005], g[2005], all;
void visit(int u, int p) {
  indexx[u] = low[u] = ++cnt;
  st.push(u);
  for (auto v : g[u]) {
    if (v == p) continue;
    if (indexx[v] != 0)
      low[u] = min(low[u], indexx[v]);
    else {
      visit(v, u);
      low[u] = min(low[u], low[v]);
    }
  }
  if (low[u] == indexx[u]) {
    bcc++;
    while (!st.empty()) {
      int x = st.top();
      st.pop();
      belong[x] = bcc;
      if (x == u) break;
    }
  }
}
void setup() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
void prepare() {
  for (int i = 1; i <= n; i++)
    if (indexx[i] == 0) visit(i, -1);
  for (int u = 1; u <= n; u++)
    for (auto v : g[u]) {
      int p = belong[u];
      int q = belong[v];
      if (p == q) continue;
      a[p].push_back(q);
    }
  for (int u = 1; u <= bcc; u++) {
    sort(a[u].begin(), a[u].end());
    a[u].resize(unique(a[u].begin(), a[u].end()) - a[u].begin());
  }
  for (int u = 1; u <= n; u++) sz[belong[u]]++;
}
void DFS(int u, int p) {
  s[u] = sz[u];
  dp[u] = 0;
  for (auto v : a[u]) {
    if (v == p) continue;
    DFS(v, u);
    dp[u] += dp[v] + s[v] * sz[u];
    s[u] += s[v];
  }
}
int getAns(int ROOT) {
  memset(dp, 0, sizeof dp);
  DFS(ROOT, -1);
  all.clear();
  for (auto u : a[ROOT]) all.push_back(s[u]);
  for (int i = 0; i <= all.size(); i++)
    for (int j = 0; j <= n; j++) f[i][j] = 0;
  f[0][0] = 1;
  for (int i = 0; i < all.size(); i++)
    for (int sum = 0; sum <= n; sum++) {
      f[i + 1][sum] |= f[i][sum];
      if (sum >= all[i]) f[i + 1][sum] |= f[i][sum - all[i]];
    }
  int anss = 0;
  for (int sum = 0; sum <= n; sum++)
    if (f[(int)all.size()][sum]) anss = max(anss, sum * (n - sum - sz[ROOT]));
  return anss + dp[ROOT];
}
void solve() {
  int res = 0;
  for (int i = 1; i <= bcc; i++) res += sz[i] * sz[i];
  int ans = 0;
  for (int i = 1; i <= bcc; i++) ans = max(ans, getAns(i));
  cout << res + ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  setup();
  prepare();
  solve();
}
