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 std::io::Read;
use std::collections::VecDeque;
use std::cmp::max;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let m: usize = iterator.next().unwrap().parse().unwrap();
    let mut graph: Vec<Vec<usize>> = vec![Vec::new(); n];
    for _ in 0..m {
        let ai: usize = iterator.next().unwrap().parse().unwrap();
        let bi: usize = iterator.next().unwrap().parse().unwrap();
        let ai = ai - 1;
        let bi = bi - 1;
        graph[ai].push(bi);
        graph[bi].push(ai);
    }

    let mut ans: usize = 0;
    let mut visited: Vec<bool> = vec![false; n];
    for i in 0..n {
        if visited[i] {
            continue;
        }
        let mut a: usize = 0;
        let mut q: VecDeque<usize> = VecDeque::new();
        visited[i] = true;
        q.push_back(i);
        a += 1;
        while let Some(u) = q.pop_front() {
            for &v in &graph[u] {
                if visited[v] {
                    continue;
                }
                visited[v] = true;
                q.push_back(v);
                a += 1;
            }
        }
        ans = max(ans, a);
    }
    ans.to_string()
}