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 itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};
use std::cmp::*;
use std::collections::*;
use std::iter::Iterator;

#[allow(unused_macros)]
macro_rules! max {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      max($x,max!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! min {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      min($x,min!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

#[derive(PartialEq, Debug, Clone)]
pub struct UnionFindNode {
  pub par: usize,
  pub rank: usize,
}

#[derive(PartialEq, Debug, Clone)]
pub struct UnionFind(pub Vec<UnionFindNode>);

impl UnionFind {
  //初期化
  pub fn new(size: usize) -> UnionFind {
    let mut vec = Vec::new();
    for i in 0..size {
      vec.push(UnionFindNode { par: i, rank: 0 });
    }

    UnionFind(vec)
  }

  //根を求める
  pub fn find(&mut self, x: usize) -> usize {
    if self.0[x].par == x {
      x
    } else {
      let par = self.0[x].par;
      let v = self.find(par);
      self.0[x].par = v;
      v
    }
  }

  //xとyの集合を併合
  pub fn unite(&mut self, x: usize, y: usize) {
    let x = self.find(x);
    let y = self.find(y);
    if x == y {
      return;
    }

    if self.0[x].rank < self.0[y].rank {
      self.0[x].par = y;
    } else {
      self.0[y].par = x;
      if self.0[x].rank == self.0[y].rank {
        self.0[x].rank += 1;
      }
    }
  }

  //xとyが同じ集合に属するか
  pub fn same(&mut self, x: usize, y: usize) -> bool {
    self.find(x) == self.find(y)
  }
}

fn main() {
  input! {
  n: usize,
  m: usize,
  mut friend: [(Usize1,Usize1); m],
  }
  let mut uf = UnionFind::new(n);

  for i in 0..m {
    uf.unite(friend[i].0, friend[i].1);
  }

  debug!(uf);

  let mut veccnt = vec![0;n];
  for ufvec in uf.0 {
    veccnt[ufvec.par as usize] = veccnt[ufvec.par as usize] + 1;
  }
  println!("{}", veccnt.iter().max().unwrap());
}