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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;
use proconio::marker::Usize1;
use std::collections::HashSet;

fn dfs(
  seen: &mut HashSet<usize>,
  frends: &mut HashSet<usize>,
  map: &Vec<Vec<usize>>,
  from: usize
) {
  if seen.contains(&from) { return }
  seen.insert(from);
  frends.insert(from);
  
  for to in map[from].iter() {
    dfs(seen, frends, map, *to);
  }
}

fn main() {
  input! {
    n: usize,
    m: usize,
    pairs: [(Usize1,Usize1);m]
  }
  
  let mut map: Vec<Vec<usize>> = vec![vec![];n];
  let mut seen: HashSet<usize> = HashSet::new();
  for (from, to) in pairs {
    map[from].push(to);
    map[to].push(from);
  }
  let mut max = 0;
  
  for from in 0..n {
    let mut frends = HashSet::new();
    seen.insert(from);
    frends.insert(from);
    for to in map[from].iter() {
      dfs(&mut seen, &mut frends, &map, *to);
    }
    
    max = std::cmp::max(max, frends.len());
  }
  
  println!("{}", max);
}