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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include<queue>
using namespace std;
int temp  = 0;
void printbfs(int ** edges,int n,int sv){
    queue<int> pendingvertices;
    int flag = 0;
    bool * visited = new bool [n];
    for(int i=0;i<n;i++){
        visited[i] = false;
    }
    
    
    for(int i=0;i<n;i++){
        if(!visited[i]){
            visited[i] = true;
            pendingvertices.push(i);
            while(!pendingvertices.empty()){
                int currentvertex = pendingvertices.front();
                
                pendingvertices.pop();
//                if(flag != 0){
//                    cout << currentvertex << " ";
//                }
                for(int i=0;i<n;i++){
                    if(i == currentvertex){
                        continue;
                    }
                    if(edges[currentvertex][i] == 1 && !visited[i]){
                        temp++;
                        pendingvertices.push(i);
                        
                        visited[i] = true;
                    }
                }
            }
        }
        
    }
    
    delete [] visited;
}
int main()
{
    int n,e;
    cin >> n >> e;
    int ** edges = new int * [n];
    for(int i=0;i<n;i++){
        edges[i] = new int [n];
        for(int j=0;j<n;j++){
            edges[i][j] = 0;
        }
    }
    while(e--)
    {
        int f,s;
        cin >> f >> s;
        edges[f-1][s-1] = 1;
    }
    printbfs(edges,n,0);
    cout << temp+1 << endl;
}