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::{fastout, input};

struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut parent = vec![];
        for x in 0..n {
            parent.push(x);
        }
        UnionFind {
            parent: parent,
            rank: vec![0; n],
        }
    }

    fn find(&mut self, x: usize) -> usize {
        if self.parent[x] == x {
            x
        } else {
            self.parent[x] = self.find(self.parent[x]);
            self.parent[x]
        }
    }

    fn unite(&mut self, x: usize, y: usize) {
        let x = self.find(x);
        let y = self.find(y);
        if x == y {
            return;
        }
        if self.rank[x] < self.rank[y] {
            self.parent[x] = y;
        } else {
            self.parent[y] = x;
            if self.rank[x] == self.rank[y] {
                self.rank[x] += 1;
            }
        }
    }

    fn same(&mut self, x: usize, y: usize) -> bool {
        self.find(x) == self.find(y)
    }
}

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

    let mut uf = UnionFind::new(n);
    for (a, b) in friends {
        let a = a - 1;
        let b = b - 1;
        uf.unite(a, b);
    }

    let mut groups = vec![0u64; n];
    for x in 0..n {
        let p = uf.find(x);
        groups[p] += 1;
    }

    let ans = groups.iter().max().copied().unwrap();
    println!("{}", ans)
}