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 maxn = 2020, N = 2000020;
int head[maxn], to[N << 1], nxt[N << 1];
int vis[maxn], low[maxn], rid;
int st[maxn], top;
int dfn[maxn], Sccn;
int C[maxn];
void Tarjan(const int u, const int pre = -1) {
  vis[u] = low[u] = ++rid;
  st[++top] = u;
  for (int e = head[u], v; e != -1; e = nxt[e]) {
    v = to[e];
    if (!vis[v])
      Tarjan(v, e), low[u] = min(low[u], low[v]);
    else if (pre ^ e ^ 1)
      low[u] = min(low[u], vis[v]);
  }
  if (low[u] == vis[u]) {
    ++Sccn;
    do {
      dfn[st[top]] = Sccn;
      C[Sccn]++;
    } while (st[top--] != u);
  }
}
long long size[maxn];
vector<int> adj[maxn];
int son[maxn];
long long dfs(int u, int p) {
  long long ans = 0;
  son[u] = C[u];
  for (int i = 0, v; i < ((int)((adj[u]).size())); i++) {
    v = adj[u][i];
    if (v != p) {
      ans += dfs(v, u);
      son[u] += son[v];
    }
  }
  return ans + 1ll * (son[u] - C[u]) * C[u];
}
bitset<maxn> bit;
int n, m;
long long Solve(int u) {
  long long res = 0, mx = 0, nn = n - C[u];
  bit.reset();
  bit[0] = 1;
  for (int i = 0; i < ((int)((adj[u]).size())); i++) res += dfs(adj[u][i], u);
  for (int i = 0; i < ((int)((adj[u]).size())); i++) {
    bit |= bit << son[adj[u][i]];
  }
  res += C[u] * nn;
  for (int i = 0; i <= n; i++)
    if (bit[i]) {
      mx = max(mx, 1ll * i * (nn - i));
    }
  return mx + res;
}
int u, v, E;
long long ans, res;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) head[i] = -1;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    nxt[E] = head[u];
    to[E] = v;
    head[u] = E++;
    nxt[E] = head[v];
    to[E] = u;
    head[v] = E++;
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) Tarjan(i);
  for (int i = 0; i < E; i += 2) {
    u = to[i];
    v = to[i + 1];
    u = dfn[u];
    v = dfn[v];
    if (u != v) adj[u].push_back(v), adj[v].push_back(u);
  }
  for (int i = 1; i <= Sccn; i++) {
    ans += 1ll * C[i] * C[i];
    res = max(res, Solve(i));
  }
  cout << ans + res << endl;
  return 0;
}
