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

#[fastout]
fn main() {
    input!{
        n:usize,
        m:usize,
        ab: [(Usize1, Usize1) ; m]
    }
    let mut graph = vec![vec![]; n];
    let mut seen = vec![false; n];
    for &(a, b) in &ab {
        graph[a].push(b);
        graph[b].push(a);
    }

    let mut stack = Vec::new();

    let mut max_vnum = 0;
    for i in 0..n {
        if seen[i] {
            continue;
        }
        seen[i] = true;
        let mut vnum = 1;
        stack.push(i);
        while let Some(v) = stack.pop() {
            for &next in &graph[v] { 
                if seen[next] {
                    continue;
                }
                seen[next] = true;
                stack.push(next);
                vnum += 1;
            }
        }
        max_vnum = max_vnum.max(vnum);
    }
    println!("{}", max_vnum);
}