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)

mod union_find {

///素集合データ構造
pub struct UnionFind {
    v: Vec<usize>,
    h: Vec<usize>,
}

impl UnionFind {
    ///要素数nの素集合データ構造を構築する。
    pub fn new(n: usize) -> UnionFind {
        let mut v = Vec::new();
        for i in 0..n {
            v.push(i);
        }
        let h = vec![1; n];
        UnionFind { v: v, h: h }
    }

    ///親を探索する。
    pub fn find(&mut self, a: usize) -> usize {
        if a == self.v[a] {
            return a;
        }

        let buf = self.v[a];
        let parent = self.find(buf);
        self.v[a] = parent;
        parent
    }

    ///親が同じか判定する。
    pub fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    ///結合する。
    pub fn union(&mut self, a: usize, b: usize) {
        let parent_a = self.find(a);
        let parent_b = self.find(b);
        if self.h[parent_a] < self.h[parent_b] {
            self.v[parent_a] = parent_b;
        } else {
            self.v[parent_b] = parent_a;

            if self.h[parent_a] == self.h[parent_b] {
                self.h[parent_a] += 1;
            }
        }
    }
}
}
#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

use std::collections::HashMap;
use union_find::*;

fn main() {
    input! {
        n:usize,m:usize,
    }
    let mut uf = UnionFind::new(n);
    for _ in 0..m {
        input! {
            a:Usize1,
            b:Usize1,
        }
        uf.union(a, b);
    }
    let mut map = HashMap::new();
    for i in 0..n {
        let q = uf.find(i);
        *map.entry(q).or_insert(0) += 1;
    }
    let mut ans = 0;
    for (i, x) in map {
        ans = std::cmp::max(ans, x);
    }
    println!("{}", ans);
}