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 kotlin.math.max

fun main() {
    solve()
}

private fun solve() {

    val (N, M) = readInts()

    val unionFind = UnionFind(N+1)

    repeat(M) {
        val (A, B) = readInts()


        unionFind.union(A, B )
    }
    var ans = 0
    unionFind.size.forEach {
        ans = max(ans, it)
    }
    println(ans)

//    println("size arr "+unionFind.size.joinToString())
//    println("parent arr "+unionFind.parent.joinToString())
}


class UnionFind(private val n: Int) {
    val parent: IntArray = IntArray(n)
    val size = IntArray(n)

    init {
        parent.forEachIndexed { index, i ->
            parent[index] = index
            size[index] = 1
        }
    }

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

    fun union(a: Int, b: Int) {
        val rootA = find(a)
        val rootB = find(b)

//        println("rootA $rootA sizeA ${size[rootA]} rootB $rootB ${size[rootB]}")
        if (rootA == rootB) return

        if (size[rootA] >= size[rootB]) {
            size[rootA] += size[rootB]
            parent[rootB] = rootA
        } else {
            size[rootB] += size[rootA]
            parent[rootA] = rootB
        }

//        println("AFTER PROCESS rootA $rootA sizeA ${size[rootA]} rootB $rootB ${size[rootB]}")
    }
}

private fun readLn() = readLine()!! // string line
private fun readInt() = readLn().toInt() // single int
private fun readStrings() = readLn().split(" ") // list of strings
private fun readInts() = readStrings().map { it.toInt() } // list of ints