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 <stdio.h>

int root(int *a, int i){
	if(a[i] < 0) return i;
	else return root(a, a[i]);
}
int main(void) {
	int N, M;
	scanf("%d %d", &N, &M);
	int arr[N];
	for(int i=0; i<N; i++) arr[i] = -1;
	int a, b, ad, bd, ans = 0;
	for(int i=0; i<M; i++){
		scanf("%d %d", &a, &b);
		ad = root(arr, --a);
		bd = root(arr, --b);
//		printf("%d %d\n", ad, bd);
		if(ad < bd){
			arr[ad]--;
			arr[bd] = ad;
		}else if(ad == bd){
		}else{
			arr[bd]--;
			arr[ad] = bd;
		}
	}
	for(int i=0; i<M; i++) if(ans > arr[i]) ans = arr[i];
	printf("%d\n", -1 * ans);
	return 0;
}