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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
 
int main() {
  long long N,M;
	cin>>N>>M;
	vector<long long> cnt(N);
	vector<vector<long long>> dp(N,vector<long long>(N));
	rep(i,M){
	  long long A,B,t,s;
		cin>>A>>B;
		if(dp[A-1][B-1]==0){
		  cnt[A-1]+=1;
			cnt[B-1]+=1;
			dp[A-1][B-1]=1;
			dp[B-1][A-1]=1;
		}
	}
	long long ans=0;
	rep(i,N)if (cnt[i]>ans) ans=cnt[i];
	cout << ans+1 << "\n";
	return 0;
}