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 java.util.*
fun findRoot(group : IntArray,  id : Int) : Int {
    var begin = id
    while(group[begin - 1] != begin){
        var temp = begin
        begin = group[begin - 1]
        group[temp - 1] = group[begin - 1]
    }
    return begin
}

fun main(args: Array<String>) {
    val reader = Scanner(System.`in`)

    val N = reader.nextInt()
    val M = reader.nextInt()

    val members = IntArray(N)
    for(index in 0..N-1) {
        members[index] = index + 1
    }

    val weights = IntArray(N)
    weights.fill(1)

    var maxGroupMembers = 0
    repeat(M){
        val a = reader.nextInt()
        val b = reader.nextInt()

        val rootA = findRoot(members, a)
        //println(rootA)
        val rootB = findRoot(members, b)
        //println(rootB)

        if(rootA != rootB) {
            if(weights[rootA - 1] + weights[rootB - 1] > maxGroupMembers){
                maxGroupMembers = weights[rootA - 1] + weights[rootB - 1]
            }

            if (weights[rootA - 1] <= weights[rootB - 1]) {
                weights[rootA - 1] += weights[rootB - 1]
                members[rootB - 1] = rootA
            }
            else {
                weights[rootB - 1] += weights[rootA - 1]
                members[rootA - 1] = rootB
            }
        }
    }

    println(if(M == 0) 1 else maxGroupMembers)
}