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>
using namespace std;
#define pi acos(-1.0)
//cout << fixed << setprecision(8) << a << endl;
#define Fast_Input ios_base :: sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);
#define pb push_back
#define mem(a, b)     memset(a, b, sizeof(a))
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef long long ll;
typedef tree<int,null_type,greater<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
/*sort(v.begin(), v.end(),
     [](const pair<int, int>& x, const pair<int, int>& y)
{
    if (x.first != y.first)
        return x.first < y.first;
    return x.second < y.second;
});
*/
int const fx[]= {+1,+0,-1,+1};
int const fy[]= {-1,+1,+0,+0};
const int inf = numeric_limits<int>::max();
const int mx=2e5;
const ll mod=1e9+7;
int par[mx+5];
int fin(int r)
{
    if(par[r]==r)return r;
    par[r]=fin(par[r]);
    return par[r];
}
int main()
{
    int n,m,x,y,u,v,i;
    cin >> n >> m;
    for(i=1;i<=n;i++)par[i]=i;
    while(m--)
    {
        cin >> x >> y;
        u=fin(x);
        v=fin(y);
        if(u!=v)par[u]=v;
    }
    map<int,int>mp;
    int maxi=-1;
    for(i=1;i<=n;i++)
    {
        mp[fin(i)]++;
        if(mp[fin(i)]>maxi)
            maxi=mp[fin(i)];
    }
    cout << maxi << endl;
    return 0;
}