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: Accepted)

#include <iostream>
#include <vector>
#include <set>
#include <queue>
using namespace std;
#define rep(i, n)for(int i=0; i<n; i++)

int main(){
  int N, M;
  cin >> N >> M;
  vector<set<int> > G(N);
  int a, b;
  //グラフ作成
  rep(i, M){
    cin >> a >> b;
    G[a-1].insert(b-1);
    G[b-1].insert(a-1);
  }

  //探索済みフラグ
  bool visited[N];
  rep(i, N)
    visited[i] = false;

  int ans = 0;
  
  queue<int> q;
  rep(i, N){
    //探索済みなら飛ばす
    if(visited[i] )
      continue;

    //頂点iから始める
    q.push(i);
    visited[i] = true;

    //連結成分のサイズ
    int size_ = 1;

    while(!q.empty() ){
      //キューの先頭の頂点を探索済みにマーク
      int now = q.front();
      q.pop();

      for(auto next : G[now] ){
	//次の頂点が未探索ならキューに追加
	if(!visited[next]){
	  q.push(next);
	  visited[next] = true;
	  //連結成分の大きさを数える
	  size_++;
	}
      }
    }// while
    ans = max(ans, size_);
  }

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