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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Linq;
using System.Runtime.InteropServices.ComTypes;


class Program
{
    static void Main(string[] args)
    {
        var inp = arrInt();
        int n = inp[0];
        int m = inp[1];
        var fMap = mapInt(m);
        UnionFind uni = new UnionFind(n);
        foreach (var p in fMap) uni.unite(p[0]-1, p[1]-1);
        int max = 0;
        for (int i = 0; i < n; i++) max = Math.Max(max, uni.groupCount(i));
        Console.WriteLine(max);        
    }




    //入力受け取り用
    static string crS() { return Console.ReadLine(); }
    static int crI() { return int.Parse(Console.ReadLine()); }
    static int[] arrInt() { return Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray(); }
    static long[] arrLong() { return Console.ReadLine().Split(' ').Select(v => long.Parse(v)).ToArray(); }
    static string[] arrStr() { return Console.ReadLine().Split(' ').ToArray(); }
    static List<int[]> mapInt(int h)
    {
        List<int[]> mapInt = new List<int[]>();
        for (int i = 0; i < h; i++)
        {
            mapInt.Add(arrInt());
        }
        return mapInt;
    }

    static List<string[]> mapStr(int h)
    {
        List<string[]> mapStr = new List<string[]>();
        for (int i = 0; i < h; i++)
        {
            mapStr.Add(arrStr());
        }
        return mapStr;
    }

    static List<char[]> mapChar(int h)
    {
        List<char[]> mapChar = new List<char[]>();
        for (int i = 0; i < h; i++)
        {
            string s = crS();
            mapChar.Add(s.ToCharArray());
        }
        return mapChar;
    }

    static List<char[]> mapCharCopy(List<char[]> map)
    {
        List<char[]> copyMap = new List<char[]>();
        foreach (var arr in map)
        {
            List<char> list = new List<char>();
            foreach (var ele in arr)
            {
                list.Add(ele);
            }
            copyMap.Add(list.ToArray());
        }
        return copyMap;
    }
}

 class UnionFind
    {
        int n;
        int[] par;
        int[] rank;

        public UnionFind(int n)
        {
            this.n = n;

            par = new int[n];
            rank = new int[n];

            //n要素で初期化
            for(int i = 0; i < n; i++)
            {
                par[i] = i;
                rank[i] = 0;
            }
        }

        //木の根を求める
        public int find(int x)
        {
            if(par[x] == x)
            {
                return x;
            } else
            {
                return par[x] = find(par[x]);
            }
        }

        //xとyの属する集合を併合
        public void unite(int x, int y)
        {
            x = find(x);
            y = find(y);
            if (x == y) return;
            if(rank[x] < rank[y])
            {
                par[x] = y;
            }
            else
            {
                par[y] = x;
                if (rank[x] == rank[y]) rank[x]++;
            }
        }

        //xとyが同じ集合に属するか否か
        public bool same(int x,int y)
        {
            return find(x) == find(y);
        }

        //木の根の個数が何個あるか
        public int rootCount()
        {
            List<int> rootList = new List<int>();
            for(int i = 0; i< n; i++)
            {
                int root = this.find(i);
                if (rootList.Contains(root)) rootList.Add(root);
            }
            return rootList.Count;
        }

        //x番目のグループの要素数を返す
        public int groupCount(int x)
        {
            int count = 0;
            int xRoot = find(x);
            for (int i = 0; i < n; i++)
            {
                int root = this.find(i);
                if (xRoot == root) count++;
            }
            return count;
        }

    }