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)

#![allow(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::*;

pub struct UnionFind {
    pub root: Vec<usize>,
}
impl UnionFind {
    pub fn new(nodes: usize) -> Self {
        Self {
            root: (0..=nodes).collect(),
        }
    }
    pub fn merge(&mut self, a: usize, b: usize) {
        let a = self.root_of(a);
        let b = self.root_of(b);

        if a < b {
            self.root[a] = b;
        } else {
            self.root[b] = a;
        }
    }
    pub fn same_root(&mut self, a: usize, b: usize) -> bool {
        self.root_of(a) == self.root_of(b)
    }
    pub fn root_of(&mut self, a: usize) -> usize {
        if self.root[a] == a {
            return a;
        }
        self.root[a] = self.root_of(self.root[a]);
        self.root[a]
    }
}

fn main() {
    input! {
        n: usize,
        rel: [(Usize1, Usize1)]
    }

    let mut uf = UnionFind::new(n);

    for &(a, b) in &rel {
        uf.merge(a, b);
    }

    let mut map = std::collections::HashMap::new();

    for i in 0..n {
        map.entry(uf.root_of(i)).and_modify(|x| *x += 1).or_insert(1);
    }

    let mut ans = 0;
    for (&k, &v) in &map {
        ans = std::cmp::max(ans, v);
    }

    println!("{}", ans);
}