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: Runtime Error)

#[allow(unused_imports)]
use std::{cmp::Reverse, collections::*};

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Chars, Usize1},
};

#[allow(unused_imports)]
use itertools::Itertools;

#[allow(unused_imports)]
use num::Integer;

#[fastout]
pub fn main() {
    input! {
        _n: usize, m: usize,
        p: [(Usize1, Usize1); m],
    }
    let mut f = Vec::new();
    p.iter()
        .map(|(a, b)| if a > b { (b, a) } else { (a, b) })
        .sorted_by(|(la, lb), (ra, rb)| if la == ra { lb.cmp(&rb) } else { la.cmp(&ra) })
        .unique()
        .for_each(|(a, b)| {
            if let Some(s) = f
                .iter_mut()
                .find(|s: &&mut HashSet<_>| s.contains(&a) || s.contains(&b))
            {
                s.insert(a);
                s.insert(b);
            } else {
                let mut t = HashSet::new();
                t.insert(a);
                t.insert(b);
                f.push(t);
            }
        });
    println!("{}", f.iter().map(|s| s.len()).max().unwrap());
}