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: Runtime Error)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <utility>
#include <iostream>
#include <functional>
#include <bitset>
#include <algorithm>
#include <vector>
#include <forward_list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <tuple>
#include <numeric>
#define rep(i, s, g) for ((i) = (s); (i) < (g); ++(i))
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const ll MOD = 1e9 + 7;
const ll INF = (1ll << 60);

struct unionfind
{
    vector<long long> d;

    unionfind(long long n) : d(n, -1) {}

    long long root(long long x)
    {
        if (d[x] < 0)
        {
            return x;
        }
        else
        {
            return d[x] = root(d[x]);
        }
    }

    bool unite(long long x, long long y)
    {
        x = root(x);
        y = root(y);

        if (x == y)
        {
            return false;
        }

        if (d[x] > d[y])
        {
            swap(x, y);
        }

        d[x] += d[y];
        d[y] = x;
        return true;
    }

    bool same(long long x, long long y)
    {
        return root(x) == root(y);
    }

    long long size(long long x)
    {
        return -d[root(x)];
    }
};

int main(void)
{
    int n, m;
    cin >> n >> m;

    unionfind u(n);

    for (int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;

        if (!u.same(a, b))
        {
            u.unite(a, b);
        }
    }

    ll ans = 0;

    if (m == 0)
    {
        cout << 1 << endl;
    }
    else
    {
        for (int i = 0; i < n; i++)
        {
            ans = max(u.size(i), ans);
        }

        cout << ans << endl;
    }
}