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(dead_code)]
#![allow(unused)]
use proconio::input;

mod union_find {
    pub struct UnionFind { pub(self) list: Vec<i32> }
    impl UnionFind {
        pub fn new(size: usize) -> UnionFind { UnionFind{ list: vec![-1; size] } }
        pub fn root(&self, n: usize) -> usize { if self.list[n] < 0 { n } else { self.root(self.list[n] as usize) } }
        pub fn size_at(&self, n: usize) -> i32 { std::cmp::max(0, -self.list[n]) }
        pub fn unite(&mut self, x: usize, y: usize) {
             let (mut x_root, mut y_root) = (self.root(x), self.root(y));
              if(x_root == y_root) { return; } 
              if(self.list[x_root] > self.list[y_root]) { std::mem::swap(&mut x_root, &mut y_root); } 
              self.list[x_root] += self.list[y_root]; 
              self.list[y_root] = x_root as i32; }
        pub fn are_in_same_tree(self, x: usize, y: usize) -> bool { self.root(x) == self.root(y) }  
    }
}

fn main() {
    input!{n: usize, m: usize, ab: [[usize; 2]; m]}
    let mut uf = union_find::UnionFind::new(n);
    for item in ab {
        uf.unite(item[0] - 1, item[1] - 1);
    }
    let mut max = 0;
    for i in 0..n {
        max = std::cmp::max(max, uf.size_at(i));
    }
    println!("{}", max);
}