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)

// ABC177D Friends

#include<stdio.h>
int par[200001];//親
int siz[200001];//サイズ

void ini(int n){ //初期化
  for(int i=1;i<=n;i++){
    par[i] = i;
    siz[i] = 1;
  }
}
int root(int x){ //根を返す
  if(par[x] == x) return x;
  else return par[x] = root(par[x]);
}

void uni(int x, int y){ //くっつける
  x = root(x);
  y = root(y);
  if(x != y){
    siz[x] += siz[y];
    par[y] = x;
  }
}

int size(int x){ //サイズを返す
  return siz[root(x)];
}

int max(int x, int y){
  if(x>=y) return x;
  else return y;
}

int main(void){
  int n,m,a,b,ma=0;
  scanf("%d %d",&n,&m);
  ini(n);
  for(int i=0;i<m;i++){
    scanf("%d %d",&a,&b);
    uni(a,b);
  }
  for(int i=1;i<=n;i++){
    ma = max(ma,size(i));
  }
  printf("%d",ma);
  return 0;
}