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: Time Limit Exceeded)

private fun getInts(): IntArray {
    return reader.readLine()!!.split(" ").map {it.toInt()}.toIntArray()
}
private val reader = System.`in`.bufferedReader()

// もらいながら接続関係を作る。なぜかREし始める。
private fun solver2():Int {
    val nm =  getInts()
    val n = nm[0]
    val m = nm[1]
    val connection = Array<List<Int>>(n){listOf()}
    for (i in 0 until m) {
        val ab = getInts()
        val a = ab[0]-1
        val b = ab[1]-1
        connection[a] = connection[a].plus(b)
        connection[b] = connection[b].plus(a)
    }
    // 探索をする。-1を未訪問とする。
    val groupMap = Array(n){-1}

    fun marking(i:Int,groupId:Int) {
        // 訪問済みの場合
        if (groupMap[i] != -1) {
            return
        }
        // DFS
        groupMap[i] = groupId
        connection[i].map{
            marking(it,groupId)
        }
    }
    for (i in 0 until n) {
        // グループのIDはその中の一番若い奴
        marking(i,i)
    }
    return groupMap.groupBy { it }.map { it.value.count()}.max()!!
}

// データを一旦もらってから接続関係を作る。
private fun getData(): Pair<Int,Array<Pair<Int,Int>>>{
    val nm =  getInts()
    var ret = arrayOf<Pair<Int,Int>>()
    for (i in 0 until nm[1]){
        val ab = getInts()
        val a = ab[0]-1
        val b = ab[1]-1
        ret = ret.plus(Pair(a,b))
    }
    return Pair(nm[0],ret)
}
private fun solver(n:Int,rel: Array<Pair<Int,Int>>):Int {
    val connection = Array<MutableList<Int>>(n){ mutableListOf() }
    for (r in rel) {
        connection[r.first].add(r.second)
        connection[r.second].add(r.first)
    }
    // 探索をする。-1を未訪問とする。
    val groupMap = Array(n){-1}

    fun marking(i:Int,groupId:Int) {
        // 訪問済みの場合
        if (groupMap[i] != -1) {
            return
        }
        // DFS
        groupMap[i] = groupId
        connection[i].map{
            marking(it,groupId)
        }
    }
    for (i in 0 until n) {
        // グループのIDはその中の一番若い奴
        marking(i,i)
    }
    return groupMap.groupBy { it }.map { it.value.count()}.max()!!
}

fun main(){
    // print(solver2())
    val d = getData()
    print(solver(d.first,d.second))
}