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: Wrong Answer)

use proconio::input;

fn main() {
    input! {
        n: usize,
        m: usize,
        pairs: [(usize, usize); m],
    }

    let mut friendship = vec![None; n+1];
    let mut group_sizes = vec![0; n+1];
    let mut id = 0;

    for p in &pairs {
        if friendship[p.0] == None && friendship[p.1] == None {
            //make new group
            id+=1;
            friendship[p.0] = Some(id);
            friendship[p.1] = Some(id);
            group_sizes[id] = 2;
        } else if friendship[p.0] == None {
            friendship[p.0] = friendship[p.1];
            group_sizes[friendship[p.1].unwrap()]+=1;
        } else if friendship[p.1] == None {
            friendship[p.1] = friendship[p.0];
            group_sizes[friendship[p.0].unwrap()]+=1;
        }
    }

    let ans = if m == 0 {
        n
    } else {
        *group_sizes.iter().max().unwrap()
    };
    println!("{}", ans);
}