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>
#define pb push_back
#define sf scanf
#define pf printf
using namespace std;
#define ll long long
const int N=2e6+9;
bool vis[N];
vector<int>adj[N];
int n,m,cnt;
void dfs(int s)
{
	vis[s]=1;
	cnt ++;
	for(auto v : adj[s]) {
		if(!vis[v]) dfs(v);
	}
}
int main()
{
    cin >> n >> m;
	int ans=0;
	while (m--)
	{
		int u,v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	for(int i=1;i<=n;i++) {
		if(!vis[i]) {
			cnt = 0;
			dfs(i);
			ans=max(ans,cnt);
		}
	}
	cout<<ans<<endl;
    return 0;
}