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)

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

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

  pub struct UnionFind {
    n: usize,
    root: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}
 
impl UnionFind {
    pub fn new(n: usize) -> Self {
        let root = (0..n).collect();
        let rank = vec![0;n];
        let size = vec![1;n];
        Self {
            n,
            root,
            rank,
            size,
        }
    }
 
    pub fn find(&mut self, x: usize) -> usize {
        if x >= self.n {
            panic!();
        } else if self.root[x] == x {
            x
        } else {
            let root = self.find(self.root[x]);
            self.root[x] = root;
            root
        }
    }
 
    pub fn unite(&mut self, x: usize, y: usize) {
        if x >= self.n || y >= self.n {
            panic!();
        }
        let x_root = self.find(x);
        let y_root = self.find(y);
        if x_root == y_root {
            return;
        }
 
        //yに所属が移る
        if self.rank[x_root] < self.rank[y_root] {
            self.root[x_root] = y_root;
            self.size[y_root] += self.size[x_root];
        } else {
            //xに所属が移る
            self.root[y_root] = x_root;
            self.size[x_root] += self.size[y_root];
            //木の高さが同じとき，片方がもう片方のroot直下につながるので1だけ高くなる
            if self.rank[x_root] == self.rank[y_root] {
                self.rank[x_root] += 1;
            }
        }
    }
 
    pub fn size(&mut self, x: usize) -> usize {
        if x >= self.n {
            panic!();
        }
        let x_root = self.find(x);
        self.size[x_root]
    }
 
    pub fn is_same(&mut self, x: usize, y: usize) -> bool{
        if x >= self.n || y >= self.n {
            panic!();
        }
        self.find(x) == self.find(y)
    }
 
}

#[fastout]
fn main() {
    input!{
        n: usize,
        m: usize,
        ab: [[usize; 2]; m],
    }

    let mut friend_relation = UnionFind::new(n);
    let mut friends = vec![0;n];
    for i in ab {
        friend_relation.unite(i[0]-1, i[1]-1);
        friends[i[0] - 1] += 1;
        friends[i[1] - 1] += 1;
    }

    let mut ans = 0;
    for i in 0..n {
        ans = max(ans, friend_relation.size(i));
    }

    println!("{}", ans);

}