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, marker::Usize1};

fn main() {
    input!(n: usize, friendship: [(Usize1, Usize1)]);

    let friendship = {
        let mut adj = vec![vec![]; n];
        for (a, b) in friendship {
            adj[a].push(b);
            adj[b].push(a);
        }
        adj
    };
    let mut gr = Graph {
        adj: &friendship,
        visited: &mut vec![false; n],
    };
    let ans = (0..n).filter_map(|v| dfs(&mut gr, v)).max().unwrap();

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

fn dfs(g: &mut Graph, v: usize) -> Option<usize> {
    if g.visited[v] {
        return None;
    }
    g.visited[v] = true;
    let adj: usize = g.adj[v].iter().filter_map(|&w| dfs(g, w)).sum();
    Some(adj + 1)
}

struct Graph<'a> {
    adj: &'a Vec<Vec<usize>>,
    visited: &'a mut Vec<bool>,
}