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)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {
        n: usize,
        m: usize,
        mut relations: [(usize, usize); m],
    }

    let mut union_found = vec![0; n + 1];
    for item in relations {
        let (mut x, mut y) = item;
        while union_found[x] != 0 {
            x = union_found[x];
        }
        while union_found[y] != 0 {
            y = union_found[y];
        }
        if x == y {
            continue;
        } else if x < y {
            union_found[y] = x;
        } else {
            union_found[x] = y;
        }
    }

    let mut count = vec![0; n + 1];
    for i in 0..n + 1 {
        let mut tmp = i;
        while union_found[tmp] != 0 {
            tmp = union_found[tmp];
        }
        count[tmp] += 1;
    }

    count.sort();
    println!("{}", count[count.len() - 1]);
}