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 = 2000 + 77;
int n, m, gp[N], t = 2, dis[N], top[N], sz[N], sum[N], A, ts;
vector<int> a[N], b[N];
bool M[N];
bitset<N> T;
void dfs(int v, int prev) {
  M[v] = 1;
  top[v] = dis[v];
  for (int u : a[v]) {
    if (u == prev) {
      continue;
    }
    if (M[u]) {
      top[v] = min(top[v], top[u]);
      continue;
    }
    dis[u] = dis[v] + 1;
    dfs(u, v);
    top[v] = min(top[v], top[u]);
  }
}
void dfs_calc(int v) {
  M[v] = 1;
  for (int u : a[v]) {
    if (M[u]) {
      continue;
    }
    if (top[u] == dis[u]) {
      gp[u] = t;
      t++;
    } else {
      gp[u] = gp[v];
    }
    dfs_calc(u);
  }
}
void dfs_down(int v, int prev) {
  for (int u : b[v]) {
    if (u == prev) {
      continue;
    }
    dfs_down(u, v);
    ts += sum[u] * sz[v];
    sum[v] += sum[u];
  }
  sum[v] += sz[v];
}
int Calc(int c) {
  ts = 0;
  memset(sum, 0, sizeof(sum));
  for (int i = 0; i <= n; i++) {
    T[i] = 0;
  }
  T[0] = 1;
  dfs_down(c, c);
  for (int u : b[c]) {
    T = T | (T << sum[u]);
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    if (T[i] == 0) {
      continue;
    }
    ans = max(ans, i * (n - i - sz[c]));
  }
  ans += ts;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int v, u;
    cin >> v >> u;
    a[v].push_back(u);
    a[u].push_back(v);
  }
  dfs(1, 1);
  memset(M, 0, sizeof(M));
  gp[1] = 1;
  dfs_calc(1);
  for (int i = 1; i <= n; i++) {
    sz[gp[i]]++;
    for (int u : a[i]) {
      if (gp[u] == gp[i]) {
        continue;
      }
      b[gp[u]].push_back(gp[i]);
    }
  }
  for (int i = 1; i < t; i++) {
    sort(b[i].begin(), b[i].end());
    b[i].resize(unique(b[i].begin(), b[i].end()) - b[i].begin());
  }
  for (int i = 1; i < t; i++) {
    A = max(A, Calc(i));
  }
  for (int i = 1; i < t; i++) {
    A += sz[i] * sz[i];
  }
  cout << A;
  return 0;
}
