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 <bits/stdc++.h>

using namespace std;
const int nax = 2e5 + 5;
vector<int> g[nax];
bool visited[nax];

void dfs(int u, int& tmp) {
  if(visited[u]) {
    return;
  }
  visited[u]=1;
  tmp++;
  for(int v : g[u]) {
    dfs(v, tmp);
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, e;
  cin >> n >> e;
  while(e--) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int ans = 0;
  for(int i = 1; i <= n; i++) {
    if(!visited[i]) {
      int tmp = 0;
      dfs(i, tmp);
      ans = max(ans, tmp);
    }
  }
  cout<<ans<<'\n';
  return 0;
}