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;

struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        let mut parent = vec![0; n];
        (0..n).for_each(|i| parent[i] = i);
        let rank = vec![0; n];
        UnionFind { parent, rank }
    }

    fn root(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            self.parent[x] = self.root(self.parent[x]);
            self.parent[x]
        }
    }

    fn same(&mut self, x: usize, y: usize) -> bool {
        self.root(x) == self.root(y)
    }

    fn merge(&mut self, x: usize, y: usize) {
        let x = self.parent[x];
        let y = self.parent[y];

        if x == y {
            return;
        }

        if self.rank[x] < self.rank[y] {
            self.parent[x] = y;
        } else {
            self.parent[y] = x;
            if self.rank[x] == self.rank[y] {
                self.rank[x] += 1;
            }
        }
    }

    fn each_size(&self) -> Vec<usize> {
        use std::collections::HashMap;
        let mut map: HashMap<usize, usize> = HashMap::new();
        for i in self.parent.iter() {
            let count = map.entry(*i).or_insert(0);
            *count += 1;
        }
        let v = map.iter().map(|(_, &b)| b).collect();
        v
    }
}

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

    let mut union = UnionFind::new(n);
    for _ in 0..m {
        input! {
            x: usize,
            y: usize
        }
        union.merge(x - 1, y - 1);
    }
    println!("{}", union.each_size().iter().max().unwrap());
}