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;
namespace FGF {
const int N = 2005;
vector<int> g[N], G[N];
int n, m, dfn[N], low[N], st[N], bel[N], scc, tp, ans, siz[N], w[N], num;
bitset<N> f;
void tarjan(int u, int f) {
  dfn[u] = low[u] = ++num, st[++tp] = u;
  for (auto v : g[u])
    if (v != f) {
      if (!dfn[v])
        tarjan(v, u), low[u] = min(low[u], low[v]);
      else
        low[u] = min(low[u], dfn[v]);
    }
  if (dfn[u] == low[u]) {
    int x;
    scc++;
    do {
      x = st[tp--];
      bel[x] = scc, w[scc]++;
    } while (x != u);
  }
}
int dfs(int u, int f) {
  int sum = 0;
  siz[u] = w[u];
  for (auto v : G[u])
    if (v != f) sum += dfs(v, u), siz[u] += siz[v];
  return sum + siz[u] * w[u];
}
void work() {
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v; i <= m; i++)
    scanf("%d%d", &u, &v), g[u].push_back(v), g[v].push_back(u);
  tarjan(1, 0);
  for (int u = 1; u <= n; u++)
    for (auto v : g[u])
      if (bel[v] != bel[u]) G[bel[v]].push_back(bel[u]);
  for (int i = 1; i <= scc; i++) {
    int x = dfs(i, 0);
    f.reset(), f[0] = 1;
    for (auto v : G[i]) f |= f << siz[v];
    for (int j = 0; j < n; j++)
      if (f[j]) ans = max(ans, x + j * (n - j - w[i]));
  }
  printf("%d\n", ans);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}
