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::input;

fn main() {
    input! {
        n: usize,
        m: usize,
        ab: [(usize, usize); m],
    }
    let mut forest = UnionFind::new(n);
    for (a, b) in ab {
        forest.union(a.min(b) - 1, a.max(b) - 1);
    }
    println!(
        "{}",
        (0..n)
            .filter(|&i| forest.is_root(i))
            .map(|i| forest.size(i))
            .max()
            .unwrap_or(0)
    );
}

struct UnionFind {
    parents: Vec<usize>,
    children: Vec<Vec<usize>>,
}
impl UnionFind {
    fn new(n: usize) -> Self {
        UnionFind {
            parents: (0..n).collect(),
            children: (0..n).map(|_| Vec::new()).collect(),
        }
    }

    fn union(&mut self, primary: usize, standby: usize) -> usize {
        let primary_root = self.find(primary);
        let standby_root = self.find(standby);
        if primary_root != standby_root {
            self.parents[standby_root] = primary_root;
            self.children[primary_root].push(standby_root);
        }
        primary_root
    }

    fn find(&self, x: usize) -> usize {
        if self.is_root(x) {
            x
        } else {
            self.find(self.parents[x])
        }
    }

    fn is_root(&self, x: usize) -> bool {
        self.parents[x] == x
    }

    fn size(&self, x: usize) -> usize {
        1 + self.children[x]
            .iter()
            .fold(0, |sum, &c| sum + self.size(c))
    }
}