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>
const int maxn = 2005;
const int inf = 1e9;
using namespace std;
vector<pair<int, int>> G[maxn];
vector<int> C[maxn], v;
bool critical[maxn * maxn];
vector<int> seen, sum, col, depth, up;
void dfs(int node, int ind) {
  seen[node] = 1;
  int minFromSons = inf;
  up[node] = depth[node];
  for (auto tmp : G[node]) {
    if (seen[tmp.first] && tmp.second != ind)
      up[node] = min(up[node], depth[tmp.first]);
    else if (!seen[tmp.first]) {
      depth[tmp.first] = depth[node] + 1;
      dfs(tmp.first, tmp.second);
      minFromSons = min(minFromSons, up[tmp.first]);
    }
  }
  up[node] = min(up[node], minFromSons);
  if (minFromSons >= depth[node] && ind != -1) {
    critical[ind] = 1;
  }
}
void dfsColor(int node, int color) {
  col[node] = color;
  v[color]++;
  for (auto tmp : G[node]) {
    if (critical[tmp.second] == 1 && col[tmp.first] == 0) continue;
    if (critical[tmp.second] == 1 && col[tmp.first]) {
      C[col[tmp.first]].push_back(color);
      C[color].push_back(col[tmp.first]);
      continue;
    } else if (col[tmp.first] == 0)
      dfsColor(tmp.first, color);
  }
}
void dfsSub(int node) {
  seen[node] = 1;
  sum[node] = v[node];
  for (auto tmp : C[node])
    if (!seen[tmp]) {
      dfsSub(tmp);
      sum[node] += sum[tmp];
    }
}
int main() {
  ios_base ::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(make_pair(b, i));
    G[b].push_back(make_pair(a, i));
  }
  seen = vector<int>(n + 1, 0);
  depth = vector<int>(n + 1, 0);
  up = vector<int>(n + 1, 0);
  dfs(1, -1);
  int c = 0;
  col = vector<int>(n + 1, 0);
  v = vector<int>(n + 1, 0);
  int all = 0;
  for (int i = 1; i <= n; ++i)
    if (col[i] == 0) {
      ++c;
      dfsColor(i, c);
      all += v[c] * v[c];
    }
  int ans = 0;
  for (int i = 1; i <= c; ++i) {
    seen = vector<int>(c + 1, 0);
    sum = vector<int>(c + 1, 0);
    dfsSub(i);
    int tmp = 0;
    for (int j = 1; j <= c; ++j) tmp += sum[j] * v[j];
    vector<int> can(n + 1, 0);
    can[0] = 1;
    for (auto tmp : C[i]) {
      for (int j = n; j >= 0; --j)
        if (can[j] && j + sum[tmp] <= n) can[j + sum[tmp]] = 1;
    }
    int res = -1;
    for (int j = (n - v[i]) / 2; j >= 0; --j)
      if (can[j]) {
        res = j;
        break;
      }
    tmp += res * (n - v[i] - res);
    ans = max(ans, tmp);
  }
  cerr << all << "\n";
  cout << ans << "\n";
}
