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: Time Limit Exceeded)

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

use proconio::input;

// NOT WORK
// TLE はともかく WA になる理由はまったくわからない
// せめてテストケース

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

    if m == 0 {
        println!("1");
        return;
    }

    let mut grp = 0;
    let mut vgrp = vec![0; n + 1];
    let mut vparent = vec![];
    for i in 0..n + 1 {
        vparent.push(i);
    }
    // let idx = vparent.clone();
    let mut vgrpnum = vec![0; n + 1];
    for _ in 0..m {
        input! {
            a: usize,
            b: usize,
        }
        // println!("a, b: {}, {}", a, b);
        if vgrp[a] == 0 && vgrp[b] == 0 {
            // new
            grp += 1;
            vgrp[a] = grp;
            vgrp[b] = grp;
            vgrpnum[grp as usize] = 2;
        } else if vparent[vgrp[a]] != 0 && vparent[vgrp[b]] != 0 && vparent[vgrp[a]] != vparent[vgrp[b]] {
            // merge b to a
            // update number of group members
            vgrpnum[vparent[vgrp[a]]] += vgrpnum[vparent[vgrp[b]]];
            vgrpnum[vparent[vgrp[b]]] = 0;
            // update parent
            // O(n^2): TLE
            let old_idx = vparent[vparent[vgrp[b]]];
            for j in 1..vparent.len() {
                if vparent[j] == old_idx {
                    vparent[j] = vparent[vgrp[a]];
                }
            }
        } else if vgrp[a] == 0 {
            // add a to group b
            vgrp[a] = vparent[vgrp[b]];
            vgrpnum[vparent[vgrp[b]]] += 1;
        } else if vgrp[b] == 0 {
            // add b to group a
            vgrp[b] = vparent[vgrp[a]];
            vgrpnum[vparent[vgrp[a]]] += 1;
        } else if vparent[vgrp[a]] == vparent[vgrp[b]] {
            // println!("already same group");
        }
        // println!("idx:     {:2?}", idx);
        // println!("vgrp:    {:2?}", vgrp);
        // println!("vparent: {:2?}", vparent);
        // println!("vgrpnum: {:2?}\n", vgrpnum);
    }

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