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,n) for(int i=0;i<(int)(n);++i)

class UF{
	vector<int> prsz;// parent & size
	public:
	UF(int n): prsz(n, -1){}
	int operator[](int v){return prsz[v] < 0? v: prsz[v]=operator[](prsz[v]);}//root
	bool operator()(int v, int u){//unite
		v = operator[](v); u = operator[](u);
		if(v == u) return false;
		if(size(v) < size(u)) swap(v, u);
		prsz[v] += prsz[u];
		prsz[u] = v;
		return true;
	}
	int size(int x){return -prsz[operator[](x)];}
};

int main(){
	int N, M; cin>>N>>M;
	UF uf(N);
	rep(i, M){
		int a,b; cin>>a>>b;
		uf(a, b); // unite(a,b)
	}
	int ans = 0;
	rep(i, N) ans = max(ans, uf.size(i));
	cout<< ans <<endl;
}