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, C, ans, T;
int a[N][N], b[N][N];
int dfn[N], low[N], vis[N];
int S[N], G[N], f[N], be[N];
void tarjan(int x, int fa) {
  dfn[x] = low[x] = ++T;
  vis[x] = 1;
  S[++*S] = x;
  for (int i = 1; i <= n; i++)
    if (a[x][i] && i != fa) {
      if (!dfn[i]) {
        tarjan(i, x);
        low[x] = min(low[x], low[i]);
      } else if (vis[i])
        low[x] = min(low[x], dfn[i]);
    }
  if (dfn[x] == low[x]) {
    C++;
    do {
      be[S[*S]] = C;
      G[C]++;
      vis[S[*S]] = 0;
    } while (S[(*S)--] != x);
  }
}
int fa[N];
void dfs1(int x, int fa) {
  for (int i = 1; i <= C; i++)
    if (b[x][i] && i != fa) {
      ::fa[i] = x;
      dfs1(i, x);
      S[x] += S[i] + G[i];
      f[x] += f[i];
    }
  f[x] += G[x] * S[x];
}
void dfs2(int x, int fa) {
  for (int i = 1; i <= C; i++)
    if (b[x][i] && i != fa) {
      f[i] = f[x] - G[x] * (S[i] + G[i]) + G[i] * (n - S[i] - G[i]);
      dfs2(i, x);
    }
}
bool v[N];
int work(int x) {
  memset(v, 0, sizeof(v));
  v[0] = 1;
  for (int i = 1; i <= C; i++)
    if (b[x][i]) {
      int cnt = (i == fa[x] ? n - G[x] - S[x] : G[i] + S[i]);
      for (int wzp = n - G[x]; wzp >= cnt; wzp--) v[wzp] |= v[wzp - cnt];
    }
  int ans = 0;
  for (int i = 0; i <= n - G[x]; i++)
    if (v[i]) ans = max(ans, i * (n - G[x] - i));
  return ans + f[x];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x][y] = a[y][x] = 1;
  }
  tarjan(1, 0);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i][j] && be[i] != be[j]) b[be[i]][be[j]] = 1;
  memset(S, 0, sizeof(S));
  dfs1(be[1], 0);
  dfs2(be[1], 0);
  for (int i = 1; i <= C; i++) ans = max(ans, work(i));
  for (int i = 1; i <= C; i++) ans += G[i] * G[i];
  printf("%d", ans);
}
