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)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn main() {
    input! {
        n: usize,
        m: usize,
    }

    let mut grp = 0;
    let mut vn = vec![0; n + 1];
    let mut vgrpnum = vec![0; n + 1];
    for _ in 0..m {
        input! {
            a: usize,
            b: usize,
        }
        // println!("vn: {:?}", vn);
        // println!("vgrpnum: {:?}", vgrpnum);
        // println!("a, b: {} {}", a, b);
        if vn[a] == 0 && vn[b] == 0 {
            // new
            grp += 1;
            vn[a] = grp;
            vn[b] = grp;
            vgrpnum[grp as usize] = 2;
        } else if vn[a] != 0 && vn[b] != 0 && vn[a] != vn[b] {
            // merge
            let old_grp = vn[b];
            vgrpnum[vn[a] as usize] += vgrpnum[vn[b]];
            vgrpnum[vn[b] as usize] = 0;
            for j in 0..vn.len() {
                // println!("j: {} vn[j]: {}", j, vn[j]);
                if vn[j] == old_grp {
                    vn[j] = vn[a];
                }
            }
        } else if vn[a] == 0 {
            vn[a] = vn[b];
            vgrpnum[vn[b] as usize] += 1;
        } else if vn[b] == 0 {
            vn[b] = vn[a];
            vgrpnum[vn[a] as usize] += 1;
        }
        // println!("vn: {:?}", vn);
        // println!("vgrpnum: {:?}\n", vgrpnum);
    }
    // println!("vn: {:?}", vn);
    // println!("vgrpnum: {:?}", vgrpnum);

    let mut ans = 0;
    for i in 0..vgrpnum.len() {
        ans = ans.max(vgrpnum[i]);
    }
    println!("{}", ans);
}