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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

typedef struct vector2 {
  int x, y;
}vector2;

int compare(const void *a, const void *b) {
  return ((vector2*)a)->x - ((vector2*)b)->x;
}

int main(void) {
  int n, m;
  scanf("%d%d", &n, &m);
  vector2 d[m];
  int r[n];
  for(int i = 0;i < m;i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    if(a < b) {
      d[i] = (vector2) { a, b };
    } else {
      d[i] = (vector2) { b, a };
    }
  }
  for(int i = 0;i < n;i++) r[i] = 0;
  qsort(d, m, sizeof(vector2), compare);
  r[d[0].x]++, r[d[0].y]++;
  for(int i = 1;i < m;i++) {
    if(d[i].y == d[i-1].y && d[i].x == d[i-1].x) continue;
    r[d[i].x]++, r[d[i].y]++;
  }
  // for(int i = 0;i < m;i++) {
  //   printf("%d %d\n", d[i].x, d[i].y);
  // }
  int result = 0;
  for(int i = 0;i < n;i++) {
    //printf("%d ", r[i]);
    if(result < r[i]) result = r[i];
  }

  printf("%d\n", result+1);
  return 0;
}