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: Wrong Answer)

use proconio::{input, marker::Usize1};

fn main() {
    input! {
        n: usize,
        m: usize,
    }

    let mut ff = vec![false; n];
    let mut f = vec![vec![]; n];
    for _ in 0..m {
        input!{ a: Usize1, b: Usize1 }
        f[a].push(b);
        f[b].push(a);
        ff[a] = true;
        ff[b] = true;
    }

    let mut mm = 0;
    for i in 0..n {
        if ff[i] {
            ff[i] = false;
            let mut g = 1;
            let mut q = vec![i];
            while let Some(qi) = q.pop() {
                for &j in &f[qi] {
                    if ff[j] {
                        ff[j] = false;
                        g += 1;
                        q.push(j);
                    }
                }
            }
            mm = mm.max(g);
        }
    }
    println!("{}", mm);
}