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: Time Limit Exceeded)

use proconio::input;
use std::cmp::max;

#[derive(Debug)]
struct UnionFind {
    // 自身が親ならばその集合に属する頂点数に -1 をかけたもの
    // そうでなければ親のid
    rank: Vec<i32>,
}

impl UnionFind {
    pub fn new(n: usize) -> UnionFind {
        UnionFind {
            // 各根は最初それぞれ独立なため -1 で初期化される
            rank: vec![-1; n],
        }
    }

    // 根を求める
    fn root(&mut self, x: usize) -> usize {
        if self.rank[x] < 0 { return x; }
        return self.root(self.rank[x] as usize);
    }

    fn unite(&mut self, x: usize, y: usize) -> bool {
        // 各木の根を求める
        let x_root = self.root(x);
        let y_root = self.root(y);
        // 根が一致するならもうすでに数えられているため計算しないfalse
        if x_root == y_root {
            return false;
        }
        // 二つの根を合成する
        // すでに多くの子を持っているものをxとしておく
        // のちにxを根として合成していく
        if self.rank[x_root] > self.rank[y_root] {
            self.rank.swap(x_root, y_root);
        }
        // 各根の保持するノード数を合計してして数を保存
        self.rank[x_root] += self.rank[y_root];
        // yのルートを更新
        self.rank[y_root] = x_root as i32;
        true
    }

    fn size(&mut self, x: usize) -> i32 {
        let root = self.root(x);
        -1 * self.rank[root]
    }
}

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

    let mut uf = UnionFind::new(n);
    for pair in ab {
        uf.unite(pair.0 - 1, pair.1 - 1);
    }

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

    println!("{:?}", max_size);
}