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<iostream>
#include<bits/stdc++.h>
#include<climits>

#define forn(x,n,s) for(int i = x; i < n; i += s )
#define forr(x,n,s) for(int i = x; i>=n; i -= s)

using namespace std;

void go(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  #ifndef ONLINE_JUDGE
       freopen("input.txt","r",stdin);
       freopen("output.txt","w",stdout);
  #endif
}

int main(){
    go();

    // code here
    int n,m;cin>>n>>m;
    unordered_map<int,vector<int>> List;
    forn(0,m,1){
        int a1,a2;cin>>a1>>a2;
        List[a1].push_back(a2);
        List[a2].push_back(a1);
    }
    unordered_set<int> visited;
    stack<int> st;
    int ans = n;
    for(auto node:List){
        if(visited.find(node.first)==visited.end()){
            int cnt = 0;
            st.push(node.first);
            visited.insert(node.first);
            while(!st.empty()){
                cnt++;
                int pnt = st.top();
                st.pop();
                for(auto nei:List[pnt]){
                    if(visited.find(nei)==visited.end()){
                        visited.insert(nei);
                        st.push(nei);
                    }
                }
            }
            ans = max(ans,cnt);
        }
    }
    ans = ans==1 ? 0 : ans;
    cout<<ans;

}