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: Runtime Error)

#![allow(unused_imports)]

use text_io::*;
use proconio::*;
use std::collections::*;
use std::process::exit;
use std::cmp::*;
use itertools::Itertools;

fn main() {
    let mut n=read!();
    let mut m=read!();
    let mut va = Vec::new();
    for i in 0..n {
        va.push(i);
    }
    let mut vb = [0;10];
    for i in 0..m {
        let mut c:usize = read!();
        c-=1;
        let mut d:usize = read!();
        d-=1;
        if va[c]>va[d] {
            va[c]=va[d];
        }
        else {
            va[d]=va[c];
        }
    }
    let mut g=0;
    for i in 0..n {
        vb[va[i]]+=1;
    }
    for i in 0..n {
        g=max(g,vb[i]);
    }
    println!("{}",g);
    // println!("{:?}",va);
    // println!("{:?}",vb);
}