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)

// -*- coding:utf-8-unix -*-

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {
        n: usize,
        m: usize,
        ab: [(usize, usize); m],
    }
    let mut tomodachi = EquivalenceRelation::new(n);
    let mut group_size = vec![1; n];
    let mut group_size_max = 1;
    for &(ai, bi) in &ab {
        if tomodachi.are_equivalent(ai - 1, bi - 1) {
            continue;
        }
        let groupsizea = group_size[tomodachi.find(ai - 1)];
        let groupsizeb = group_size[tomodachi.find(bi - 1)];
        tomodachi.make_equivalent(ai - 1, bi - 1);
        let newval = groupsizea + groupsizeb;
        group_size[tomodachi.find(ai - 1)] = newval;
        group_size_max = group_size_max.max(newval);
    }
    println!("{}", group_size_max);
}

use std::cell::Cell;

#[derive(Debug, Clone)]
struct EquivalenceRelation {
    parent: Vec<Cell<usize>>,
    rank: Vec<Cell<usize>>,
}

impl EquivalenceRelation {
    fn new(n: usize) -> Self {
        let mut parent = Vec::with_capacity(n);
        for i in 0..n {
            parent.push(Cell::new(i));
        }
        let rank = vec![Cell::new(0); n];
        return Self { parent, rank };
    }

    fn make_equivalent(&mut self, a: usize, b: usize) {
        let volume = self.parent.len();
        if a >= volume || b >= volume {
            panic!(
                "Tried to make {} and {} equivalent but there are only {} elements",
                a, b, volume
            );
        }
        let aa = self.find(a);
        let bb = self.find(b);
        if aa == bb {
            return;
        }
        let aarank = self.rank[aa].get();
        let bbrank = self.rank[bb].get();
        if aarank > bbrank {
            self.parent[bb].set(aa);
        // self.rank[aa] = aarank.max(bbrank + 1);
        } else {
            self.parent[aa].set(bb);
            self.rank[bb].set(bbrank.max(aarank + 1));
        }
    }

    fn find(&self, a: usize) -> usize {
        let volume = self.parent.len();
        if a >= volume {
            panic!("Tried to find {} but there are only {} elements", a, volume);
        }
        let b = self.parent[a].get();
        if b == a {
            return a;
        } else {
            let c = self.find(b);
            self.parent[a].set(c);
            return c;
        }
    }

    fn are_equivalent(&self, a: usize, b: usize) -> bool {
        return self.find(a) == self.find(b);
    }
}