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>
struct gedge {
  int to, nex;
  bool bri;
} ge[4000010];
struct tedge {
  int to, nex;
} te[4010];
int gh[2010], th[2010], s[2010], siz[2010], dfn[2010], low[2010], col[2010],
    q[2010], f[2010], tot, m;
void gadd(int a, int b) {
  tot++;
  ge[tot].to = b;
  ge[tot].nex = gh[a];
  gh[a] = tot;
  ge[tot + m].to = a;
  ge[tot + m].nex = gh[b];
  gh[b] = tot + m;
}
void tadd(int a, int b) {
  tot++;
  te[tot].to = b;
  te[tot].nex = th[a];
  th[a] = tot;
}
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
void bri(int x) {
  ge[x].bri = 1;
  if (x > m)
    ge[x - m].bri = 1;
  else
    ge[x + m].bri = 1;
}
void tarjan(int fa, int x) {
  tot++;
  dfn[x] = low[x] = tot;
  for (int i = gh[x]; i; i = ge[i].nex) {
    if (dfn[ge[i].to] == 0) {
      tarjan(x, ge[i].to);
      low[x] = min(low[x], low[ge[i].to]);
      if (low[ge[i].to] > dfn[x]) bri(i);
    } else if (dfn[ge[i].to] < dfn[x] && ge[i].to != fa)
      low[x] = min(low[x], dfn[ge[i].to]);
  }
}
void dfs(int x) {
  for (int i = gh[x]; i; i = ge[i].nex) {
    if (col[ge[i].to]) {
      if (col[ge[i].to] != col[x]) {
        tadd(col[x], col[ge[i].to]);
        tadd(col[ge[i].to], col[x]);
      }
    } else if (!ge[i].bri) {
      col[ge[i].to] = col[x];
      dfs(ge[i].to);
    }
  }
}
void dfs(int fa, int x) {
  siz[x] = s[x];
  for (int i = th[x]; i; i = te[i].nex) {
    if (te[i].to != fa) {
      dfs(x, te[i].to);
      siz[x] += siz[te[i].to];
    }
  }
}
int main() {
  int N, n, i, j, k, a, b, ans, res;
  scanf("%d%d", &n, &m);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    gadd(a, b);
  }
  tot = 0;
  tarjan(0, 1);
  N = 0;
  tot = 0;
  for (i = 1; i <= n; i++) {
    if (col[i] == 0) {
      N++;
      col[i] = N;
      dfs(i);
    }
  }
  for (i = 1; i <= n; i++) s[col[i]]++;
  ans = 0;
  for (i = 1; i <= N; i++) {
    dfs(0, i);
    res = 0;
    for (j = 1; j <= N; j++) res += s[j] * siz[j];
    q[0] = 0;
    for (j = th[i]; j; j = te[j].nex) {
      q[0]++;
      q[q[0]] = te[j].to;
    }
    memset(f, 0, sizeof(f));
    for (j = 1; j <= q[0]; j++) {
      for (k = n - s[i]; k >= siz[q[j]]; k--)
        f[k] = max(f[k], f[k - siz[q[j]]] + siz[q[j]]);
    }
    res += f[(n - s[i]) >> 1] * (n - s[i] - f[(n - s[i]) >> 1]);
    ans = max(ans, res);
  }
  printf("%d", ans);
}
