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;
use proconio::marker::Usize1;

fn find(x: usize, par: &mut Vec<usize>, rnk: &Vec<usize>) -> usize {
    //
    if par[x] == x {
        return x;
    } else {
        par[x] = find(par[par[x]], par, rnk);
        return par[x];
    }
}

fn unite(x: usize, y: usize, par: &mut Vec<usize>, rnk: &mut Vec<usize>) {
    let x = find(x, par, rnk);
    let y = find(y, par, rnk);
    if x == y {
        return;
    }

    if rnk[x] < rnk[y] {
        //
        par[x] = y;
    } else {
        par[y] = x;
        if rnk[x] == rnk[y] {
            rnk[x] += 1;
        }
    }
}

// fn same(x: usize, y: usize, par: &mut Vec<usize>, rnk: &mut Vec<usize>) -> bool {
//     find(x, par, rnk) == find(y, par, rnk)
// }

#[allow(non_snake_case)]
fn main() {
    input! {
        N: usize,
        M: usize,
        AB: [(Usize1, Usize1); M]
    }

    let mut par: Vec<_> = (0..N).collect();
    let mut rnk = vec![0; N];

    for &(a, b) in AB.iter() {
        //
        unite(a, b, &mut par, &mut rnk);
    }

    let mut cnt = vec![0; N];
    for i in 0..N {
        cnt[par[i]] += 1;
    }

    println!("{}", cnt.iter().max().unwrap());
}