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],
    }
 
    const MAX: usize = 200001;
 
    let mut bel: [usize; MAX] = [0; MAX];
    for i in 0..n {
        bel[i+1] = i+1;
    }

    let mut group: Vec<Vec<usize>> = (0..n+1).map(|x| vec![x]).collect();
 
    for (a, b) in t {
        if bel[a] != bel[b] {
            let clonedb = group[bel[b]].clone();
            for i in &group[bel[b]] { bel[*i] = bel[a]; }
            group[bel[a]].extend(clonedb);
        }
    }
    
    println!("{}", group.iter().map(|x| x.len()).max().unwrap());
}