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;
int N, M, C, D, Ans, G[2005], F[2005], S[2005], fa[2005], Bel[2005], Dfn[2005],
    Low[2005], A[2005][2005], B[2005][2005];
bool V[2005];
void Tarjan(int u, int f) {
  Dfn[u] = Low[u] = ++D;
  V[S[++*S] = u] = 1;
  for (int v = 1; v <= N; v++)
    if (A[u][v] && v != f) {
      if (!Dfn[v])
        Tarjan(v, u), Low[u] = min(Low[u], Low[v]);
      else if (V[v])
        Low[u] = min(Low[u], Low[v]);
    }
  if (Dfn[u] == Low[u]) {
    C++;
    do {
      Bel[S[*S]] = C;
      G[C]++;
      V[S[*S]] = 0;
    } while (S[(*S)--] != u);
  }
}
void DFS1(int u, int f) {
  for (int v = 1; v <= C; v++)
    if (B[u][v] && v != f)
      DFS1(v, u), S[u] += S[v] + G[v], fa[v] = u, F[u] += F[v];
  F[u] += G[u] * S[u];
}
void DFS2(int u, int f) {
  for (int v = 1; v <= C; v++)
    if (B[u][v] && v != f)
      F[v] = F[u] - G[u] * (S[v] + G[v]) + G[v] * (N - G[v] - S[v]), DFS2(v, u);
}
int Work(int u) {
  memset(V, 0, sizeof(V));
  V[0] = 1;
  for (int v = 1; v <= C; v++)
    if (B[u][v]) {
      int x = (v == fa[u] ? N - G[u] - S[u] : G[v] + S[v]);
      for (int i = N - G[u]; i >= x; i--) V[i] |= V[i - x];
    }
  int ret = 0;
  for (int i = 0; i <= N - G[u]; i++)
    if (V[i]) ret = max(ret, i * (N - G[u] - i));
  return ret + F[u];
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= M; i++) {
    int u, v;
    scanf("%d%d", &u, &v), A[u][v] = A[v][u] = 1;
  }
  Tarjan(1, 0);
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= N; j++)
      if (A[i][j] && Bel[i] != Bel[j]) B[Bel[i]][Bel[j]] = 1;
  memset(S, 0, sizeof(S));
  DFS1(Bel[1], 0);
  DFS2(Bel[1], 0);
  for (int i = 1; i <= C; i++) Ans = max(Ans, Work(i));
  for (int i = 1; i <= C; i++) Ans += G[i] * G[i];
  printf("%d\n", Ans);
}
