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 warning(disable : 4996)
using namespace std;
int n, m;
vector<int> edge[3000];
set<int> cut[3000];
vector<int> tedge[3000];
int low[3000];
int check[3000];
int depth[3000];
int num[3000];
int numi;
int dat[3000];
int dab;
int d[2][3000];
int child[3000];
int tsum;
void back(int now, int parent) {
  int i, j, k;
  check[now] = 1;
  depth[now] = parent == -1 ? 0 : depth[parent] + 1;
  int mi = depth[now];
  for (auto e : edge[now]) {
    if (check[e]) {
      if (e != parent) mi = min(mi, depth[e]);
      continue;
    }
    back(e, now);
    mi = min(mi, low[e]);
  }
  low[now] = mi;
  if (mi == depth[now] && parent + 1) {
    cut[now].insert(parent);
    cut[parent].insert(now);
  }
}
void back2(int now) {
  check[now] = 1;
  num[now] = numi;
  for (auto e : edge[now]) {
    if (check[e]) continue;
    if (cut[now].count(e)) continue;
    back2(e);
  }
}
int back3(int now, int parent) {
  int sum = dat[now];
  depth[now] = depth[parent] + 1;
  for (auto e : tedge[now]) {
    if (e == parent) continue;
    int x = back3(e, now);
    sum += x;
    tsum += dat[now] * x;
  }
  return sum;
}
int main() {
  int i, j, k;
  scanf("%d%d", &n, &m);
  for (i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    edge[x].push_back(y), edge[y].push_back(x);
  }
  back(0, -1);
  memset(check, 0, sizeof(check));
  for (i = 0; i < n; i++) {
    if (check[i]) continue;
    back2(i);
    numi++;
  }
  for (i = 0; i < n; i++) dat[num[i]]++;
  for (i = 0; i < n; i++) {
    for (auto e : cut[i]) {
      tedge[num[i]].push_back(num[e]);
    }
  }
  for (i = 0; i < numi; i++) dab += dat[i] * dat[i];
  int rdab = 0;
  for (i = 0; i < numi; i++) {
    fill(d[0], d[0] + n, 0);
    fill(depth, depth + n, 0);
    d[0][0] = 1;
    tsum = 0;
    int xsum = 0;
    for (auto e : tedge[i]) {
      int x = back3(e, i);
      xsum += x;
      tsum += x * dat[i];
      for (j = 0; j < n; j++) d[1][j] = d[0][j];
      for (j = 0; j < n; j++) {
        if (d[0][j]) d[1][j + x] = 1;
      }
      for (j = 0; j < n; j++) d[0][j] = d[1][j];
    }
    int ma = 0;
    for (j = 0; j < n; j++) {
      if (d[0][j]) {
        ma = max(ma, j * (xsum - j));
      }
    }
    int sum = 0;
    rdab = max(rdab, dab + tsum + ma);
  }
  printf("%d", rdab);
  return 0;
}
