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)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use proconio::*;
use text_io::*;

const U_INF: usize = 1 << 60;
const I_INF: isize = 1 << 60;

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

    let mut uf = UnionFindTree::new(n);
    let mut vec = vec![0; n + 1];

    for (ai, bi) in ab {
        uf.unite(ai - 1, bi - 1);
    }

    for i in 0..n {
        vec[uf.root(i)] += 1;
    }

    println!("{}", vec.into_iter().max().unwrap());
}

struct UnionFindTree {
    parents: Vec<usize>,
    sizes: Vec<usize>,
    weights: Vec<isize>,
    has_loops: Vec<bool>,
}

impl UnionFindTree {
    fn new(n: usize) -> Self {
        let parents = (0..n).collect();
        let sizes = vec![1; n];
        let weights = vec![0; n];
        let has_loops = vec![false; n];
        UnionFindTree {
            parents,
            sizes,
            weights,
            has_loops,
        }
    }

    fn root(&mut self, x: usize) -> usize {
        if self.parents[x] == x {
            x
        } else {
            let tmp = self.root(self.parents[x]);
            self.weights[x] += self.weights[self.parents[x]];
            self.parents[x] = tmp;
            tmp
        }
    }

    fn size(&self, x: usize) -> usize {
        self.sizes[x]
    }

    fn has_loop(&self, x: usize) -> bool {
        self.has_loops[x]
    }

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

    fn unite_with_weight(&mut self, x: usize, y: usize, w: isize) {
        let root_x = self.root(x);
        let root_y = self.root(y);

        if self.same(x, y) {
            self.has_loops[root_x] = true;
            self.has_loops[root_y] = true;
        } else {
            if self.sizes[root_x] >= self.sizes[root_y] {
                self.parents[root_y] = root_x;
                self.sizes[root_x] += self.sizes[root_y];
                self.weights[root_y] = -w - self.weights[y] + self.weights[x];
            } else {
                self.parents[root_x] = root_y;
                self.sizes[root_y] += self.sizes[root_x];
                self.weights[root_x] = w + self.weights[y] - self.weights[x];
            }
        }
    }

    fn unite(&mut self, x: usize, y: usize) {
        self.unite_with_weight(x, y, 0);
    }

    fn diff(&mut self, x: usize, y: usize) -> Option<isize> {
        if self.same(x, y) {
            Some(self.weights[x] - self.weights[y])
        } else {
            None
        }
    }

    fn is_parent(&self, x: usize) -> bool {
        self.parents[x] == x
    }
}