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;

namespace D
{
    class Program
    {
        static void Main(string[] args)
        {
            var NM = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            var N = NM[0]; var M = NM[1];
            var group = new UnionFind(N);

            for (int i = 0; i < M; i++)
            {
                var ab = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
                group.unite(ab[0], ab[1]);
            }

            var result = 0;
            for (int i = 1; i <= N; i++)
            {
                if (group.RootCount(i) > result) result = group.RootCount(i);
            }
            Console.WriteLine(result);
        }
    }
    public class UnionFind
    {
        private int[] parents;
        // 同じ根に属する枝の数を格納する
        private int[] rootCount;

        public int RootCount(int branch)
        {
            return rootCount[root(branch)];
        }

        public UnionFind(int count)
        {
            parents = new int[count + 1];
            rootCount = new int[count + 1];
            for (int i = 0; i < count + 1; i++)
            {
                parents[i] = i;
                rootCount[i] = 1;
            }
        }

        private int root(int N)
        {
            if (parents[N] == N) return N;
            return parents[N] = root(parents[N]);
        }

        //木を併合する際に、枝の数も合算する
        public void unite(int X, int Y)
        {
            int rootX = root(X);
            int rootY = root(Y);
            if (rootX == rootY) return;
            rootCount[rootX] += rootCount[rootY];
            parents[rootY] = rootX;
        }

        public bool sameRoot(int X, int Y)
        {
            return root(X) == root(Y);
        }
    }
}