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)

import kotlin.math.*
import java.util.*

const val MOD = 1_000_000_007L

fun readI() = readLine()!!.toInt()
fun readL() = readLine()!!.toLong()
fun readD() = readLine()!!.toDouble()
fun readIs() = readLine()!!.split(' ').map(String::toInt)
fun readLs() = readLine()!!.split(' ').map(String::toLong)
fun readDs() = readLine()!!.split(' ').map(String::toDouble)

var y = 0

fun dfs(g: Map<Int, List<Int>>, i: Int, c: BooleanArray) {
  c[i] = true
  y++
  val h = g[i] ?: return
  for (e in h) {
    if (c[e]) continue
    dfs(g, e, c)
  }
}

fun main() {
  val (n, m) = readIs()
  val g = mutableMapOf<Int, MutableList<Int>>()
  repeat(m) {
    val (a, b) = readIs()
    g.computeIfAbsent(a) { mutableListOf() }.add(b)
    g.computeIfAbsent(b) { mutableListOf() }.add(a)
  }
  val c = BooleanArray(n + 1)
  var x = 0
  for (i in 1..n) {
    if (c[i]) continue
    y = 0
    dfs(g, i, c)
    x = max(x, y)
  }
  println(x)
}