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)




class UnionFind(val size: Int) {
    private val vec = IntArray(size){-1}
    fun root(a: Int): Int {
        return if (vec[a] < 0) {
            a
        }else {
            val root = root(vec[a])
            vec[a] = root
            root
        }
    }
    fun sizeOf(a: Int): Int {
        return -vec[root(a)]
    }
    fun unite(a: Int, b: Int) {
        val ar = root(a)
        val br = root(b)
        if (ar == br) return
        if (vec[ar] < vec[br]) {
            vec[ar] += vec[br]
            vec[br] = ar
        }else {
            vec[br] += vec[ar]
            vec[ar] = br
        }
    }
}
fun main() {
    val (n, m) = readLine()!!.trim().split(' ').map(String::toInt)
    val uft = UnionFind(n)
    repeat(m) {
        val (a, b) = readLine()!!.trim().split(' ').map(String::toInt)
        uft.unite(a - 1, b - 1)
    }
    println((0 until n).map{uft.sizeOf(it)}.max()!!)
}