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 std::cerr;
using std::endl;
inline int rd() {
  int a = 1, b = 0;
  char c = getchar();
  while (!isdigit(c)) a = c == '-' ? 0 : 1, c = getchar();
  while (isdigit(c)) b = b * 10 + c - '0', c = getchar();
  return a ? b : -b;
}
const int N = 2005;
int n, m;
std::vector<int> G[N], T[N];
int dfn[N], low[N], num, stk[N], top;
int id[N], cnt, w[N], vis[N];
long long size[N], dep[N];
long long ans;
void tarjan(int x, int fa) {
  dfn[x] = low[x] = ++num;
  stk[++top] = x;
  for (int y : G[x]) {
    if (y == fa) continue;
    if (!dfn[y]) {
      tarjan(y, x);
      low[x] = std::min(low[x], low[y]);
    } else
      low[x] = std::min(low[x], dfn[y]);
  }
  if (dfn[x] == low[x]) {
    int y;
    ++cnt;
    do {
      y = stk[top--];
      id[y] = cnt;
      ++w[cnt];
    } while (x != y);
  }
}
void dfs(int x, int fa) {
  size[x] = w[x], dep[x] = w[x] * w[x];
  for (int y : T[x]) {
    if (y == fa) continue;
    dfs(y, x);
    size[x] += size[y];
    dep[x] += dep[y] + size[y] * w[x];
  }
}
inline void solve(int x) {
  memset(vis, 0, sizeof(vis));
  vis[0] = 1;
  int all = 0;
  long long tmp = w[x] * w[x], res = 0;
  for (int y : T[x]) {
    dfs(y, x);
    tmp += dep[y] + size[y] * w[x];
    for (int i = all; ~i; --i)
      if (vis[i]) vis[i + size[y]] = 1;
    all += size[y];
  }
  for (int i = 0; i <= all; ++i)
    if (vis[i]) res = std::max(res, i * (n - w[x] - i) + tmp);
  ans = std::max(ans, res);
}
int main() {
  n = rd(), m = rd();
  for (int i = 1; i <= m; ++i) {
    int x = rd(), y = rd();
    G[x].push_back(y);
    G[y].push_back(x);
  }
  tarjan(1, 0);
  for (int x = 1; x <= n; ++x)
    for (int y : G[x])
      if (id[x] != id[y]) T[id[x]].push_back(id[y]);
  for (int i = 1; i <= n; ++i) solve(i);
  std::cout << ans << std::endl;
  return 0;
}
