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(non_snake_case)]
#![allow(dead_code)]
#![allow(unused_macros)]
#![allow(unused_imports)]

use std::str::FromStr;
use std::io::*;
use std::collections::*;
use std::cmp::*;

struct Scanner<I: Iterator<Item = char>> {
    iter: std::iter::Peekable<I>,
}

macro_rules! exit {
    () => {{
        exit!(0)
    }};
    ($code:expr) => {{
        if cfg!(local) {
            writeln!(std::io::stderr(), "===== Terminated =====")
                .expect("failed printing to stderr");
        }
        std::process::exit($code);
    }}
}

impl<I: Iterator<Item = char>> Scanner<I> {
    pub fn new(iter: I) -> Scanner<I> {
        Scanner {
            iter: iter.peekable(),
        }
    }

    pub fn safe_get_token(&mut self) -> Option<String> {
        let token = self.iter
            .by_ref()
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        if token.is_empty() {
            None
        } else {
            Some(token)
        }
    }

    pub fn token(&mut self) -> String {
        self.safe_get_token().unwrap_or_else(|| exit!())
    }

    pub fn get<T: FromStr>(&mut self) -> T {
        self.token().parse::<T>().unwrap_or_else(|_| exit!())
    }

    pub fn vec<T: FromStr>(&mut self, len: usize) -> Vec<T> {
        (0..len).map(|_| self.get()).collect()
    }

    pub fn mat<T: FromStr>(&mut self, row: usize, col: usize) -> Vec<Vec<T>> {
        (0..row).map(|_| self.vec(col)).collect()
    }

    pub fn char(&mut self) -> char {
        self.iter.next().unwrap_or_else(|| exit!())
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.get::<String>().chars().collect()
    }

    pub fn mat_chars(&mut self, row: usize) -> Vec<Vec<char>> {
        (0..row).map(|_| self.chars()).collect()
    }

    pub fn line(&mut self) -> String {
        if self.peek().is_some() {
            self.iter
                .by_ref()
                .take_while(|&c| !(c == '\n' || c == '\r'))
                .collect::<String>()
        } else {
            exit!();
        }
    }

    pub fn peek(&mut self) -> Option<&char> {
        self.iter.peek()
    }
}

type Graph<T> = Vec<Vec<T>>;

struct Solver {
    visited: Vec<bool>,
    cnt: usize,
}

impl Solver {
    fn dfs(&mut self, i: usize, g: &Graph<usize>) {
        if self.visited[i] {
            return;
        }
        self.visited[i] = true;
        self.cnt += 1;
        for &n in &g[i] {
            self.dfs(n, &g);
        }
    }
}

fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin.bytes().map(|c| c.unwrap() as char));
    let N: usize = sc.get();
    let M: usize = sc.get();
    let mut adj = vec![Vec::new(); N];
    for _ in 0..M {
        let A: usize = sc.get();
        let B: usize = sc.get();
        adj[A-1].push(B-1);
        adj[B-1].push(A-1);
    }
    let mut s = Solver { visited: vec![false; N], cnt: 0 };
    let mut ans = 1;
    for i in 0..N {
        let before = s.cnt;
        s.dfs(i, &adj);
        let after = s.cnt;
        ans = max(ans, after-before);
    }
    println!("{}", ans);
}