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, f[2005], v[4000000], o[4000000], p, t, d[2005], l[2005], g[2005], grp,
    w[2005], f2[2005], v2[2005 + 2005], o2[2005 + 2005], s[2005], ans, res;
bool bg[4000000], u[2005], dp[2005];
void bd(int a, int b) {
  v[++p] = b;
  o[p] = f[a];
  f[a] = p;
}
void bd2(int a, int b) {
  v2[++p] = b;
  o2[p] = f2[a];
  f2[a] = p;
}
void dfs(int x, int y) {
  l[x] = d[x] = ++t;
  for (int i = (f[x]); i; i = o[i])
    if (v[i] != y) {
      if (!d[v[i]]) {
        dfs(v[i], x);
        l[x] = min(l[x], l[v[i]]);
      } else
        l[x] = min(l[x], d[v[i]]);
      if (l[v[i]] == d[v[i]]) bg[i] = 1;
    }
}
void part(int x, int y) {
  if (!g[x]) {
    g[x] = ++grp;
    if (y) {
      bd2(g[x], g[y]);
      bd2(g[y], g[x]);
    }
  }
  w[g[x]]++;
  u[x] = 1;
  for (int i = (f[x]); i; i = o[i])
    if (!u[v[i]]) {
      if (!bg[i]) g[v[i]] = g[x];
      part(v[i], x);
    }
}
void size(int x, int y) {
  int sum = w[x] * w[x];
  res += w[x] * w[x];
  s[x] = w[x];
  for (int i = (f2[x]); i; i = o2[i])
    if (v2[i] != y) {
      size(v2[i], x);
      s[x] += s[v2[i]];
      res += w[x] * s[v2[i]];
      sum += w[x] * s[v2[i]];
    }
}
int main() {
  scanf("%d %d", &n, &m);
  while (m--) {
    int a, b;
    scanf("%d %d", &a, &b);
    bd(a, b);
    bd(b, a);
  }
  dfs(1, 0);
  p = 0;
  part(1, 0);
  for (int i = (1); i <= (grp); i++) {
    res = w[i] * n;
    for (int j = (0); j <= (n); j++) dp[j] = 0;
    dp[0] = 1;
    for (int j = (f2[i]); j; j = o2[j]) {
      size(v2[j], i);
      for (int k = (n); k >= (s[v2[j]]); k--)
        if (dp[k - s[v2[j]]]) dp[k] = 1;
    }
    for (int j = (0); j <= (n); j++)
      if (dp[j]) ans = max(ans, res + j * (n - w[i] - j));
  }
  printf("%d\n", ans);
  scanf("\n");
}
