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;

#define rep(i, a, b) for(int i=a; i<=b; i++)
#define trav(a, x) for(auto& a : x)
#define all(x) begin(x), end(x)
#define sz(x) (int) x.size()
#define pb push_back
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const int MOD = 1e9+7;

int n, m, a, b;
vi adj[200001];
bool visited[200001];
int ans=0;
int cnt=0;

void dfs(int x){
	if(visited[x]) return;
	visited[x]=true;
	cnt++;
	trav(k, adj[x]){
		if(!visited[k]){
			dfs(k);
		}
	}

}

int main(){
  // freopen("input.txt", "r", stdin); 
  // freopen("output.txt", "w", stdout);
  cin.sync_with_stdio(0); cin.tie();
  cin.exceptions(cin.failbit);

  cin >> n >> m;
  rep(i, 1, n) adj[i].clear();
  rep(i, 0, n) visited[i]=false;
  rep(i, 1, m){
  	cin >> a >> b;
  	adj[a].pb(b);
  	adj[b].pb(a);
  }

  rep(i, 1, n){
  	if(visited[i]) continue;
  	cnt=0; dfs(i);
  	ans=max(ans, cnt);
  }

  cout << ans;
}