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: Accepted)

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define input freopen("input.txt", "r", stdin);
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);



#include<bits/stdc++.h>
using namespace std;
#define maxn 200005
int n, e, par[maxn], cnt[maxn];
void make_set()
{
    for(int i=1; i<=n; i++) par[i]=i;
}
int find_rep(int u)
{
    if(par[u]==u) return u;
    return par[u] = find_rep(par[u]);
}
void make_union(int u, int v)
{
    u = find_rep(u);
    v = find_rep(v);
    if(u!=v) par[v]=u;
}
int main()
{
    cin>>n>>e;
    make_set();
    for(int i=1; i<=e; i++){
        int u, v;
        cin>>u>>v;
        make_union(u, v);
    }
    int ans=0;
    for(int i=1; i<=n; i++){
        int id = find_rep(i);
        cnt[id]++;
        ans = max(ans, cnt[id]);
    }
    cout<<ans;
}