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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {
    static int n;
    static int m;
    static int[] ai;
    static int[] bi;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        m = in.nextInt();
        ABC177DUnionFind unionFind = new ABC177DUnionFind(n);
        ai = new int[m + 1];
        bi = new int[m + 1];
        int[] directFriendCount = new int[n+1];
        for (int i = 1; i <= m; i++) {
            ai[i] = in.nextInt();
            bi[i] = in.nextInt();
            unionFind.unite(ai[i], bi[i]);
            directFriendCount[ai[i]]++;
            directFriendCount[bi[i]]++;
        }
        
        Queue<Integer> remainQueue = new LinkedList<>();
        for(int i=1; i<=n; i++) {
            if(directFriendCount[i] == 0) {
                continue;
            }
            remainQueue.add(i);
        }
        
        int maxGroupSize = 1;
        while(!remainQueue.isEmpty()) {
            Queue<Integer> nextQueue = new LinkedList<>();
            int a = remainQueue.poll();
            int groupSize = 1;
            while(!remainQueue.isEmpty()) {
                int b = remainQueue.poll();
                if(unionFind.isInSameGroup(a, b)) {
                    groupSize++;
                }
                else {
                    nextQueue.add(b);
                }
            }
            maxGroupSize = Math.max(maxGroupSize, groupSize);
            remainQueue = nextQueue;
        }
        
        System.out.println(maxGroupSize);
    }
}
class ABC177DUnionFind {
    private int[] parent;
    public ABC177DUnionFind(int n) {
        this.parent = new int[n+1];
        for(int i=1; i<=n; i++) {
            this.parent[i] = i; // みんな、自分がroot
        }
    }
    public int root(int a) {
        if(this.parent[a] == a) {
            return a;
        }
        this.parent[a] = this.root(this.parent[a]);
        return this.parent[a];
    }
    public void unite(int a, int b) {
        int rootA = this.root(a);
        int rootB = this.root(b);
        if(rootA == rootB) {
            return;
        }
        this.parent[rootA] = rootB;
    }
    public boolean isInSameGroup(int a, int b) {
        return this.root(a) == this.root(b);
    }
}