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(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num_traits::clamp;
use ordered_float::OrderedFloat;
use proconio::{input, marker::*, fastout};
use superslice::*;

#[fastout]
fn main() {
    input! {
        n: usize, m: usize,
        ab: [(Usize1, Usize1); m],
    }

    let mut uft = UFT::new(n);
    for (a, b) in ab {
        uft.merge(a, b);
    }

    let mut rank = vec![0usize; n];
    for i in 0..n {
        rank[uft.root(i)] += 1;
    }
    let ans = rank.iter().max().unwrap();
    println!("{}", ans);
}

#[allow(dead_code)]
/// Union Find Tree
pub struct UFT {
    pub par: Vec<usize>,
    pub rank: Vec<usize>,
}
impl UFT {
    #[allow(dead_code)]
    pub fn new(n: usize) -> Self {
        UFT {
            par: (0..n).collect(),
            rank: vec![0; n],
        }
    }
    #[allow(dead_code)]
    pub fn root(&mut self, x: usize) -> usize {
        if self.par[x] == x {
            x
        } else {
            let p = self.par[x];
            let pp = self.root(p);
            self.par[x] = pp;
            pp
        }
    }
    #[allow(dead_code)]
    pub fn merge(&mut self, x: usize, y: usize) {
        let x = self.root(x);
        let y = self.root(y);
        if x == y {
            return;
        }
        if self.rank[x] < self.rank[y] {
            self.par[x] = y;
        } else {
            self.par[y] = x;
            if self.rank[x] == self.rank[y] {
                self.rank[x] += 1;
            }
        }
    }
}