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: Runtime Error)

fun main(args: Array<String>) {
    val (n, m) = readLine()!!.split(' ').map { it.toInt() }

    val ufs = UFS(n)
    for (i in 0 until m) {
        val arr = readLine()!!.split(' ').map { it.toInt() }
        ufs.union(arr[0] - 1, arr[1] - 1)
    }

    val map = HashMap<Int, Int>()
    for (i in 0 until n) {
        val key = ufs.find(i)
        map[key] = map.getOrDefault(key, 0) + 1
    }
    var ans = 0
    map.forEach { (key, value) ->
        ans = maxOf(ans, value)
    }
    println(ans)
}

class UFS(var n: Int = 0) {
    private val parent = IntArray(n) { i -> i }
    private val rank = IntArray(n)

    fun find(x: Int): Int {
        if (x != parent[x]) {
            parent[x] = find(parent[x])
        }
        return parent[x]
    }

    fun union(x: Int, y: Int): Boolean {
        val px = find(x)
        val py = find(y)
        if (px == py) {
            return false
        }
        when {
            rank[px] > rank[py] -> parent[py] = px
            rank[px] < rank[py] -> parent[px] = py
            else -> {
                parent[px] = py
                rank[px]++
            }
        }
        return true
    }
}