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: Wrong Answer)

fun main(args: Array<String>) {
    val (n, m) = getInts()

    val a = IntArray(n) { -1 }
    repeat(m) {
        val (x, y) = getInts().map { it - 1 }.sorted()
        a[y] = x
    }

    val k = HashMap<Int, Int>()
    a.indices.forEach {
        val p = parentOf(it, a)
        k[p] = (k[p] ?: 0) + 1
    }
    println(k.map { it.value }.max())
}

fun parentOf(k: Int, a: IntArray) : Int {
    if (a[k] == -1) return k
    else return parentOf(a[k], a).also {
        a[k] = it
    }
}

fun getInt() = readLine()!!.toInt()
fun getInts() = readLine()!!.split(" ").map { it.toInt() }