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)

import kotlin.math.*
import java.io.BufferedReader
import java.io.InputStreamReader
import java.util.*


private var fs = BufferedReader(InputStreamReader(System.`in`))
private fun readLn() = fs.readLine()!! // string line
private fun readStrings() = readLn().split(" ") // list of strings
private fun readLong() = readLn().toLong()
private fun readLongs() = readStrings().map { it.toLong() } // list of ints
private fun readInt() = readLn().toInt() // single int
private fun readInts() = readStrings().map { it.toInt() } // list of ints

fun main() {
    solveAtc()
}
fun solveAtc() {
    var (n,m) = readInts()
    var a = ArrayList<Pair<Int,Int>>()
    for (i in 0 until m) {
        var (x,y) = readInts()
        if(!a.contains(Pair(min(x,y),max(x,y))))
            a.add(Pair(min(x,y),max(x,y)))
    }
    var map = HashMap<Int,Int>()
    for (i in 0 until a.size) {
        if(!map.contains(a[i].first)) map[a[i].first] = 1
        else{
            map[a[i].first] = map.getValue(a[i].first) + 1
        }
    }
    var ans = 0
    for(i in map.entries) ans = max(ans,i.value)
    println(ans +1)
}