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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
vector<int> mp[2005];
int dfn[2005], sta[2005], low[2005], color, ii, cnt, belong[2005];
bool ins[2005];
void tarjan(int x, int p) {
  dfn[x] = low[x] = ++ii;
  sta[++cnt] = x;
  ins[x] = 1;
  for (auto child : mp[x])
    if (child != p) {
      if (!dfn[child]) {
        tarjan(child, x);
        low[x] = min(low[x], low[child]);
      } else if (ins[child]) {
        low[x] = min(low[x], dfn[child]);
      }
    }
  if (dfn[x] == low[x]) {
    ++color;
    int t;
    do {
      t = sta[cnt];
      cnt--;
      ins[t] = 0;
      belong[t] = color;
    } while (dfn[t] != low[t]);
  }
}
int sz[2005];
set<int> mp1[2005];
int tt;
bitset<2005> bs;
int dfs(int x, int pa) {
  int ans = 0;
  for (int c : mp1[x])
    if (c != pa) {
      ans += dfs(c, x);
    }
  tt += ans * sz[x];
  return ans + sz[x];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, u, v; i < m; i++) {
    scanf("%d%d", &u, &v);
    mp[u].push_back(v);
    mp[v].push_back(u);
  }
  tarjan(1, 0);
  for (int(u) = 1; (u) <= (int)(n); (u)++) {
    sz[belong[u]]++;
    for (int v : mp[u])
      if (belong[u] != belong[v]) {
        mp1[belong[u]].insert(belong[v]);
        mp1[belong[v]].insert(belong[u]);
      }
  }
  int ans = 0;
  for (int(i) = 1; (i) <= (int)(color); (i)++) {
    tt = 0;
    bs.reset();
    bs[0] = 1;
    for (int c : mp1[i]) {
      int x = dfs(c, i);
      bs = bs | (bs << x);
    }
    tt += sz[i] * (n - sz[i]);
    int ma = 0;
    for (int(j) = 1; (j) <= (int)(2005 - 1); (j)++)
      if (bs[j] == 1) {
        ma = max(ma, j * (n - sz[i] - j));
      }
    tt += ma;
    ans = max(tt, ans);
  }
  for (int(i) = 1; (i) <= (int)(color); (i)++) ans += sz[i] * sz[i];
  printf("%d\n", ans);
  return 0;
}
