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;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
const int N = 2e3 + 7, M = 1e6 + 7;
int n, m, h[N], cnt = 1, col[N], val[N], u[M], v[M], z[N];
bool vis[N], mark[M];
vector<pair<int, int> > g[N], d;
vector<int> t[N];
void cut(int x, int pr = -1) {
  vis[x] = true;
  for (auto i : g[x])
    if (i.second != pr) {
      if (vis[i.first]) {
        while (d.size() && h[d.back().second] >= h[i.first]) d.pop_back();
      } else {
        h[i.first] = h[x] + 1;
        d.push_back({i.second, x});
        cut(i.first, i.second);
        if (d.size() && d.back().second == x)
          mark[i.second] = true, d.pop_back();
      }
    }
}
int prep(int x, int c) {
  col[x] = c;
  int ans = 1;
  for (auto i : g[x])
    if (!col[i.first] && !mark[i.second]) ans += prep(i.first, c);
  return ans;
}
int pap;
int dfs(int x, int pr = -1) {
  z[x] = 0;
  for (auto i : t[x])
    if (i != pr) z[x] += dfs(i, x);
  if (pr != -1) pap += val[x] * z[x];
  z[x] += val[x];
  return z[x];
}
int calc(int x) {
  pap = 0;
  dfs(x);
  bitset<N> b;
  b[0] = 1;
  for (auto i : t[x]) b = b | (b << z[i]);
  int mx = 0;
  for (int i = 0; i < N; i++)
    if (b[i]) mx = max(mx, ((n - i - val[x]) * i));
  return mx + pap + (n - val[x]) * val[x];
}
int32_t main() {
  n = in(), m = in();
  int ans = 0;
  for (int i = 0; i < m; i++) {
    int x = in() - 1, y = in() - 1;
    u[i] = x, v[i] = y;
    g[x].push_back({y, i});
    g[y].push_back({x, i});
  }
  cut(0);
  for (int i = 0; i < n; i++)
    if (!col[i]) val[cnt] = prep(i, cnt), ans += val[cnt] * val[cnt], cnt++;
  for (int i = 0; i < m; i++)
    if (mark[i])
      t[col[u[i]]].push_back(col[v[i]]), t[col[v[i]]].push_back(col[u[i]]);
  int an = 0;
  for (int i = 1; i < cnt; i++) an = max(an, calc(i));
  cout << an + ans << endl;
}
