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 petgraph::unionfind::UnionFind;
use proconio::fastout;
use proconio::input;
use proconio::marker::Usize1;
use std::collections::HashMap;

#[fastout]
fn main() {
    input! {
        n: usize, m: usize,
        abs: [(Usize1, Usize1); m],
    }
    let mut uf: UnionFind<_> = UnionFind::new(n);
    for &(a, b) in &abs {
        uf.union(a, b);
    }
    let mut counts = HashMap::new();
    for i in 0..n {
        let x = uf.find(i);
        *counts.entry(x).or_insert(0) += 1;
    }
    println!("{}", counts.values().max().unwrap());
}