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)

struct UnionFind {
    roots: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut roots = vec![];
        for i in 0..n {
            roots.push(i);
        }
        UnionFind { roots: roots }
    }
    fn find(&mut self, i: usize) -> usize {
        if self.roots[i] == i {
            i
        } else {
            let tmp_i = self.roots[i];
            self.roots[i] = self.find(tmp_i);
            self.roots[i]
        }
    }
    fn unite(&mut self, i: usize, j: usize) {
        let root_i = self.find(i);
        let root_j = self.find(j);
        self.roots[root_i] = root_j;
    }
}

fn main() {
    proconio::input! {
        n: usize,
        m: usize,
        abs: [(usize, usize); m],
    }
    let abs = abs
        .iter()
        .cloned()
        .map(|(a, b)| (a - 1, b - 1))
        .collect::<Vec<_>>();
    let mut union_find = UnionFind::new(n);
    for (a, b) in abs {
        union_find.unite(a, b);
    }
    let mut m = std::collections::HashMap::new();
    (0..n).map(|i| union_find.find(i)).for_each(|x| {
        *m.entry(x).or_insert(0) += 1;
    });
    let ans = m.values().max().unwrap();
    println!("{}", ans);
}