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;
#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars};
#[allow(unused_imports)]
use std::cmp::{min, max};

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


	let mut gnext = vec![0usize; n];	// 人xが所属する現在グループのリンク
	for i in 0..n {
		gnext[i] = i;
	}

	let mut gnum = vec![1usize; n];	// 人xが所属する現在グループのリンク

	for i in 0..m {
		let mut ga = ab[i].0-1;
		loop {
			if gnext[ga] == ga {
				break;
			}
			ga = gnext[ga];
		}
		gnext[ab[i].0-1] = ga;
		let mut gb = ab[i].1-1;
		loop {
			if gnext[gb] == gb {
				break;
			}
			gb = gnext[gb];
		}
		gnext[ab[i].1-1] = gb;
		if ga == gb {
			continue;
		}
		gnext[gb] = ga;
		gnum[ga] += gnum[gb];
		gnum[gb] = 0;
	}

	let mut ans = 0;
	for i in 0..n {
		ans = max(ans, gnum[i]);
	}
	println!("{}", ans);
}