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: Accepted)

use proconio::marker::Usize1;
fn main() {
    proconio::input! {
        n: usize,
        m: usize,
        ab: [(Usize1,Usize1);m],
    }
    let mut uf=vec![-1;n];
    for (a,b) in ab {
        unite(&mut uf, a, b);
    }
    println!("{}",(0..n).map(|i|count(&mut uf,i)).max().unwrap());
}
fn find(uf: &mut Vec<i32>, i: usize) -> usize {
    if uf[i]<0 {
        i
    }
    else {
        uf[i]=find(uf, uf[i]as usize)as i32;
        uf[i]as usize
    }
}
fn unite(uf: &mut Vec<i32>, i: usize, j: usize) {
    let mut i=find(uf, i);
    let mut j=find(uf, j);
    if i==j {
        return;
    }
    if uf[i]>uf[j] {
        std::mem::swap(&mut i, &mut j);
    }
    uf[i]+=uf[j];
    uf[j]=i as i32;
}
fn count(uf: &mut Vec<i32>, i: usize) -> usize {
    let i=find(uf, i);
    (-uf[i])as usize
}