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: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var scanner = bufio.NewScanner(os.Stdin)

func NextStr() string {
	scanner.Scan()
	return scanner.Text()
}

func NextInt() int {
	i, _ := strconv.Atoi(NextStr())
	return i
}

func max(x, y int) int {
	if x > y {
		return x
	}
	return y
}

func sort(x, y int) (int, int) {
	if x > y {
		return y, x
	}
	return x, y
}

func main() {
	scanner.Split(bufio.ScanWords)
	n, m := NextInt(), NextInt()
	groups := make([]int, n)
	belongings := make([]int, n)
	members := make([][]int, n)
	// 初期は各メンバーは一人グループ
	for i := 0; i < n; i++ {
		groups[i] = 1
		belongings[i] = i
		members[i] = []int{i}
	}

	var ans int

	for i := 0; i < m; i++ {
		a, b := NextInt()-1, NextInt()-1
		a, b = sort(a, b)
		belong := belongings[a]
		// aの所属グループにbのメンバーを所属させる
		groups[belong] += groups[b]
		target := belongings[b]
		if target != belong {
			oldmembers := members[target]
			members[belong] = append(members[belong], members[target]...)
			groups[target] = 0
			members[target] = []int{}
			// 旧メンバーの所属先をaの所属先に更新
			for _, member := range oldmembers {
				belongings[member] = belong
			}
			ans = max(ans, groups[belong])
		}
	}
	fmt.Println(ans)
}