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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using static System.Console;

class Program
{
    static void Main(string[] args)
    {
        var NM = ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
        var (N, M) = (NM[0], NM[1]);
        var AB = Enumerable.Range(0, M).Select(i => ReadLine().Split(' ')).Select(s => (A: int.Parse(s[0]) - 1, B: int.Parse(s[1]) - 1)).ToArray();

        var uf = new UnionFind(N);

        foreach (var (A, B) in AB)
            uf.Unite(A, B);

        var ans = Enumerable.Range(0, N).Select(i => uf.Size(i)).Max();

        WriteLine(ans);
    }
}
class UnionFind
{
    private int[] d;
    public UnionFind(int n) => d = Enumerable.Repeat(-1, n).ToArray();
    public int Find(int x) => (d[x] < 0) ? x : d[x] = Find(d[x]);
    public bool Unite(int x, int y)
    {
        (x, y) = (Find(x), Find(y));
        if (x == y)
            return false;
        if (d[x] > d[y])
            (x, y) = (y, x);
        d[x] += d[y];
        d[y] = x;
        return true;
    }
    public bool Same(int x, int y) => Find(x) == Find(y);
    public int Size(int x) => -d[Find(x)];
}