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: Runtime Error)

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

int main() {
  int N, M;
  cin >> N >> M;

  queue<P> q;
  rep(i,0,M-1) {
    int A, B;
    cin >> A >> B;
    q.push(P(A, B));
  }

  vector<set<int>> v(1, set<int>());
  P p = q.front(); q.pop();
  v[0].insert(p.first); v[0].insert(p.second);

  while (!q.empty()) {
    P p = q.front(); q.pop();
    bool found = false;
    for (set<int> & s : v) {
      if (s.find(p.first) != s.end() || s.find(p.second) != s.end()) {
        s.insert(p.first); s.insert(p.second);
        found = true;
      }
    }
    if (!found) {
      set<int> s;
      s.insert(p.first); s.insert(p.second);
      v.push_back(s);
    }
  }

  size_t ans = 0;
  for (set<int> s : v) {
    ans = max(ans, s.size());
  }

  cout << ans << endl;
  return 0;
}