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 main(){
    int n,m;
    scanf("%d %d",&n,&m);

    int a[n][n];
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)a[i][j]=0;

    for(int i=0;i<m;i++){
        int p,q;
        scanf("%d%d",&p,&q);
        a[p][q]=1;
        a[q][p]=1;

     for(int i=0;i<n;i++)for(int j=0;j<n;j++)if(a[i][p]==1&&a[j][q]==1){
         a[i][j]=1;
         a[j][i]=1;
     }

    }

    int max=0;

    for(int i=0;i<n;i++){
        int count=0;
        for(int j=0;j<n;j++){
            count+=a[i][j];
        }
        if(count>max)max=count;
    }

    printf("%d",max+1);


    

}