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 Kotlin (1.3.71) (judge verdict: Accepted)

import java.lang.IllegalArgumentException

fun main(args: Array<String>) {
	val (n, m) = readLine()?.split(" ")?.map {it.toInt()} ?: throw IllegalArgumentException()
	
	val friendsUnion = FriendsUnion(n)
	repeat(m) {
		val (a, b) = readLine()?.split(" ")?.map { it.toInt() } ?: throw IllegalArgumentException()
		with(friendsUnion) { addFriends(a, b) }
	}
	println(friendsUnion.maxUnionNumber())
}

data class FriendsUnion(val n: Int, val rootList: MutableList<Int> = MutableList(n+1) {it}, val unionNumberList: MutableList<Int> = MutableList(n+1) {1}) {
	fun addFriends(friend1: Int, friend2: Int) {
		//添え字が若い方を親とする
		var (parent, child) = if (friend1 < friend2) Pair(friend1, friend2) else Pair(friend2, friend1)
		//それぞれのrootを探す
		val parentRootIndex = findRoot(parent)
		val childRootIndex = findRoot(child)
		//rootが同じならそのまま
		return if (parentRootIndex == childRootIndex) return else {
			//rootが違うなら子グループを親グループにマージ。
			unionNumberList[parentRootIndex] += unionNumberList[childRootIndex]
			unionNumberList[childRootIndex] = 0
			//子グループのrootの親を、親グループのrootにしておく。
			rootList[childRootIndex] = parentRootIndex
		}
	}
	
	private fun findRoot(friend: Int): Int {
		//rootが自分だった場合、自分を返す。
		return if (rootList[friend] == friend) friend else {
			//rootが自分じゃない場合、親がrootか判定した上で、親をrootにする(後々のfindRootを早くするため)。
			rootList[friend] = findRoot(rootList[friend])
			rootList[friend]
		}
	}
	
	fun maxUnionNumber(): Int {
		return unionNumberList.max() ?: throw IllegalArgumentException()
	}
}