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: Time Limit Exceeded)

use proconio::{input};

fn main() {
    input! {
        n: usize,
        m: usize,
        t: [(usize, usize); m],
    }

    let mut belonging: Vec<Option<usize>> = vec!(None; n+1);
    let mut group: Vec<Vec<usize>> = Vec::new();

    for (a, b) in t {
        match (belonging[a], belonging[b]) {
            (None, None) => {
                let groupnum = group.len();
                group.push(vec![a, b]);
                belonging[a] = Some(groupnum);
                belonging[b] = Some(groupnum);
            },
            (Some(ia), None) => {
                belonging[b] = Some(ia);
                group[ia].push(b);
            },
            (None, Some(ib)) => {
                belonging[a] = Some(ib);
                group[ib].push(a);
            },
            (Some(ia), Some(ib)) if ia != ib => {
                for i in &group[ib] {
                    belonging[*i] = belonging[a];
                }
                let mut clonedb = group[ib].clone();
                group[ia].append(&mut clonedb);
            },
            _ => {},
        }
    }
    
    let answer = group.iter().map(|x| x.len()).max();
    match answer {
        Some(a) => println!("{}", a),
        None => println!("{}", 1),
    }
}