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 N = 2005;
int n, m, p, __time, low[N], num[N], st[N], del[N], res[N];
int cntComponents, Component[N], Weight[N], subt[N];
vector<int> child[N];
void pre_dfs(int u, int x) {
  low[u] = num[u] = ++__time;
  st[++p] = u;
  for (int v : child[u])
    if (v != x) {
      if (num[v] != 0) {
        if (!del[v]) low[u] = min(low[u], num[v]);
        continue;
      }
      pre_dfs(v, u);
      low[u] = min(low[u], low[v]);
    }
  if (low[u] >= num[u]) {
    ++cntComponents;
    while (st[p] != u) del[st[p]] = 1, Component[st[p--]] = cntComponents;
    del[st[p]] = 1, Component[st[p--]] = cntComponents;
  }
}
void dfs(int u, int x) {
  subt[u] = Weight[u];
  for (int v : child[u])
    if (v != x) dfs(v, u), subt[u] += subt[v];
  res[u] = subt[u] * Weight[u];
  for (int v : child[u])
    if (v != x) res[u] += res[v];
}
int solve(int source) {
  dfs(source, source);
  vector<int> G(n + 1);
  G[0] = 1;
  for (int v : child[source])
    for (int i = n; i >= subt[v]; --i) G[i] |= G[i - subt[v]];
  int Ans = 0;
  for (int i = 0; i <= n; ++i)
    if (G[i] != 0) Ans = max(Ans, i * (n - Weight[source] - i));
  return Ans + res[source];
}
int main() {
  scanf("%d%d", &n, &m);
  vector<pair<int, int> > Edges(m);
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    Edges[i] = make_pair(u, v);
    child[u].push_back(v), child[v].push_back(u);
  }
  pre_dfs(1, 1);
  for (int i = 1; i <= n; ++i) {
    ++Weight[Component[i]];
    child[i].clear();
  }
  for (int i = 0; i < m; ++i) {
    int a = Edges[i].first, b = Edges[i].second;
    a = Component[a], b = Component[b];
    if (a != b) child[a].push_back(b), child[b].push_back(a);
  }
  int Ans = 0;
  for (int i = 1; i <= cntComponents; ++i) Ans = max(Ans, solve(i));
  printf("%d\n", Ans);
}
