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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAX 200001
int p[MAX], sz[MAX];

void init(int N) {
  for(int i = 1; i <= N; i++) {
    p[i] = i;
    sz[i] = 1;
  }
}

//encuentra al padre que los reprsenta
int Find(int x) {
  if(p[x] == x) return x;
  sz[x] = sz[p[x]];
  //return p[x] = Find(p[x]);
  return Find(p[x]);
}

void Union(int x, int y) {
  int pX = Find(x);
  int pY = Find(y);
  p[pX] = pY;
  if(pX != pY) {
    sz[pY] = sz[pX] + sz[pY];
    sz[pX] = sz[pY];
  }
}

void print(int n) {
  for(int i = 1; i <= n; i++) {
    cout << p[i] << " ";
  }
  cout << endl;
}

void printSize(int n) {
  for(int i = 1; i <= n; i++) {
    cout << sz[i] << " ";
  }
  cout << endl;
}

int nroComponentes(int n) {
  map<int, int> m;
  int c = 0;
  for(int i = 1; i <= n; i++) {
    int parent = Find(i);
    m[parent]++;
    if(m[parent] == 1) {
      c++;
    } 
  }
  return c;
}


int nroDeNodos(int x, int n) {
  int c = 0;
  int pN = Find(x);
  for(int i = 1; i <= n; i++) {
    int parent = Find(i);
    if(parent == pN) {
      c++;
    }
  }
  return c;
}

int main() {
  int n, m;
  cin >> n >> m;
  init(n);
  /*for(int i = 1; i <= n; i++) {
    cout << i << " ";
  }
  cout << endl << endl;
  print(n);*/
  while(m--) {
    int x, y;
    cin >> x >> y;
    Union(x, y);
    //print(n);
  }
  int ans = 0;
  for(int i = 1; i <= n; i++) {
    //ans = max(ans, nroDeNodos(i, n));
    ans =  max(ans, sz[i]);
  }

  //printSize(n);
  cout << ans;
  /*cout << nroComponentes(n) << "\n";
  cout << nroDeNodos(4, n);*/
}