Problem:
Score : 400 points

Problem Statement

There are N persons called Person 1 through Person N.

You are given M facts that "Person A_i and Person B_i are friends." The same fact may be given multiple times.

If X and Y are friends, and Y and Z are friends, then X and Z are also friends. There is no friendship that cannot be derived from the M given facts.

Takahashi the evil wants to divide the N persons into some number of groups so that every person has no friend in his/her group.

At least how many groups does he need to make?

Constraints

2 \leq N \leq 2\times 10^5

0 \leq M \leq 2\times 10^5

1\leq A_i,B_i\leq N

A_i \neq B_i

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
\vdots
A_M B_M

Output

Print the answer.

Sample Input 1

5 3
1 2
3 4
5 1

Sample Output 1

3

Dividing them into three groups such as \{1,3\}, \{2,4\}, and \{5\} achieves the goal.

Sample Input 2

4 10
1 2
2 1
1 2
2 1
1 2
1 3
1 4
2 3
2 4
3 4

Sample Output 2

4

Sample Input 3

10 4
3 1
4 1
5 9
2 6

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main() {
  int n, m;
  cin >> n >> m;
  unordered_map<int, unordered_set<int> > groups;
  unordered_map<int, int> ids;
  int nextid = 1;

  unordered_map<int, int> alias;

  rep(i, m) {
    int a, b;
    cin >> a >> b;

    if (ids[a] > 0 && ids[b] > 0 && ids[a] != ids[b]) {
      vector<int> p;
      alias[ids[a]] = ids[b];
    } else if (ids[a] > 0) {
      ids[b] = ids[a];
      groups[ids[a]].insert(b);
    } else if (ids[b] > 0) {
      ids[a] = ids[b];
      groups[ids[b]].insert(a);
    } else {
      ids[a] = nextid;
      ids[b] = nextid;
      groups[nextid].insert(a);
      groups[nextid].insert(b);
      nextid++;
    }
  }

  int res = 1;

  unordered_map<int, int> counts;

  for (auto g : groups) {
    counts[g.first] = (int)g.second.size();
  }

  for (auto p : alias) {
    counts[p.first] += counts[p.second];
  }

  for (auto c : counts) {
    // printf("group: %d\n", g.first);
    // for (auto e : g.second) cout << e << endl;
    res = max(res, c.second);
  }

  cout << res << endl;

  return 0;
}