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>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int MAX = 2007;
const int MOD = 1000000007;
vector<int> g[MAX];
bool used[MAX];
int tin[MAX];
int fup[MAX];
int T = 0;
vector<int> G[MAX];
vector<pair<int, int> > e;
set<pair<int, int> > S;
void dfs(int v, int p) {
  used[v] = 1;
  tin[v] = fup[v] = T++;
  for (int i = (0); i < (g[v].size()); ++i) {
    int to = g[v][i];
    if (to == p) continue;
    if (used[to]) {
      fup[v] = min(fup[v], tin[to]);
    } else {
      dfs(to, v);
      fup[v] = min(fup[v], fup[to]);
      if (fup[to] > tin[v]) {
        e.push_back(make_pair(to, v));
        S.insert(make_pair(to, v));
        S.insert(make_pair(v, to));
      }
    }
  }
}
int cmp = 0;
int cnt[MAX];
int c[MAX];
void dfs2(int v) {
  cnt[cmp]++;
  c[v] = cmp;
  used[v] = 1;
  for (int i = (0); i < (g[v].size()); ++i) {
    if (!used[g[v][i]] && S.find(make_pair(v, g[v][i])) == S.end()) {
      dfs2(g[v][i]);
    }
  }
}
int r = 0;
int d[MAX];
int dfs3(int v) {
  used[v] = 1;
  d[v] = cnt[v];
  int t = 0;
  for (int i = (0); i < (G[v].size()); ++i) {
    if (!used[G[v][i]]) {
      t += dfs3(G[v][i]);
      d[v] += d[G[v][i]];
      t += cnt[v] * d[G[v][i]];
    }
  }
  return t;
}
bool dp[MAX];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = (0); i < (m); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1);
  memset(used, 0, sizeof(used));
  long long res = 0;
  for (int i = (0); i < (n); ++i) {
    if (!used[i]) {
      dfs2(i);
      res += cnt[cmp] * cnt[cmp];
      cmp++;
    }
  }
  for (int i = (0); i < (e.size()); ++i) {
    int a = e[i].first;
    int b = e[i].second;
    G[c[a]].push_back(c[b]);
    G[c[b]].push_back(c[a]);
  }
  int R = 0;
  for (int v = (0); v < (cmp); ++v) {
    memset(used, 0, sizeof(used));
    memset(dp, 0, sizeof(dp));
    memset(d, 0, sizeof(d));
    used[v] = 1;
    vector<int> a;
    int rr = 0;
    for (int i = (0); i < (G[v].size()); ++i) {
      rr += dfs3(G[v][i]);
      a.push_back(d[G[v][i]]);
    }
    rr += cnt[v] * (n - cnt[v]);
    dp[0] = 1;
    for (int i = (0); i < (a.size()); ++i) {
      for (int j = (MAX)-1; j >= (0); --j) {
        if (dp[j]) {
          dp[j + a[i]] = 1;
        }
      }
    }
    int r = 0;
    for (int i = (0); i < (MAX); ++i) {
      if (dp[i]) {
        r = max(r, i * (n - cnt[v] - i));
      }
    }
    rr += r;
    R = max(R, rr);
  }
  cout << res + R << endl;
  return 0;
}
