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: Time Limit Exceeded)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define PI 3.141592653589793
#define MOD 1000000007
//#define DEBUG

typedef struct HUMAN{
	struct HUMAN *friends[200100];
	int yet;
} human;

//qsort用、昇順
int compare_int(const void *a, const void *b)
{
	return *(int*)a-*(int*)b;
}
//qsort(array, 10, sizeof(int), compare_int)

int n, m;
human *man[200100];

int main(){
	scanf("%d%d", &n, &m);
	for(int i=0; i<n; i++){
		man[i] = (human *) malloc(sizeof(human));
		man[i]->yet = 0;
	}

	for(int i=0; i<m; i++){
		int a, b;
		scanf("%d%d", &a, &b);

		int j=0;
		int flug = 0;
		while(man[a-1]->friends[j] != NULL){
			if(man[a-1]->friends[j] == man[b-1]){
				flug = 1;
				break;
			}
			j++;
		}
		if(flug == 0) man[a-1]->friends[j] = man[b-1];
		j = 0;
		flug = 0;
		while(man[b-1]->friends[j] != NULL){
			if(man[b-1]->friends[j] == man[a-1]){
				flug = 1;
				break;
			}
			j++;
		}
		if(flug == 0) man[b-1]->friends[j] = man[a-1];
		
	}

	//int count = 0;
	int max = 0;
	for(int i=0; i<n; i++){
		//printf("aa\n");
		if(man[i]->yet == 0){
			//printf("h:%d\n", i);
			//count++;
			int  j=0;
			man[i]->yet = 1;
			while(man[i]->friends[j] != NULL){
				if(man[i]->friends[j]->yet == 0){
					man[i]->friends[j]->yet = 1;
					//printf("hh:%d\n", j);
				}
				j++;
			}
			if(max<j+1) max = j+1;
			//printf("j+1:%d\n", j+1);
		}
	}

	printf("%d\n", max);
	
	return 0;
}