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 Go (1.14.1) (judge verdict: Time Limit Exceeded)

package main

import (
	"fmt"
)

func main() {
	var N, M int
	fmt.Scan(&N, &M)
	uf := newUF(N)
	var x, y int
	for i := 0; i < M; i++ {
		fmt.Scan(&x, &y)
		uf.unite(x-1, y-1)
	}
	checked := make([]int, uf.n)
	for i := range checked {
		r := uf.root(i)
		checked[r]++
	}
	fmt.Println(max(checked))

}

type unionFind struct {
	pair []int
	n    int
}

func newUF(N int) *unionFind {
	uf := new(unionFind)
	uf.n = N
	pair := make([]int, N)
	for i := 0; i < N; i++ {
		pair[i] = i
	}
	uf.pair = pair
	return uf
}
func (uf unionFind) unite(x int, y int) {
	rx := uf.root(x)
	ry := uf.root(y)
	if rx != ry {
		uf.pair[rx] = ry
	}
}
func (uf unionFind) root(x int) int {
	r := x
	for {
		if uf.pair[r] == r {
			break
		}
		r = uf.pair[r]
	}
	return r
}
func (uf unionFind) same(x int, y int) bool {
	rx := uf.root(x)
	ry := uf.root(y)
	return rx == ry
}
func (uf unionFind) countTree() int {
	ans := 0
	checked := make([]bool, uf.n)
	for i := range checked {
		r := uf.root(i)
		if checked[r] {
			checked[r] = true
			ans++
		}
	}
	return ans
}
func max(a []int) int {
	max := a[0]
	for _, i := range a {
		if i > max {
			max = i
		}
	}
	return max
}