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;
using ll = long long;
#define rep(i, start, n) for (ll i = (ll)(start); i < (ll)(n); ++i)
static const ll INFTY = 1L << 62L;
ll dfs(vector<vector<ll>> &g, ll *v, ll i)
{
    v[i] = true;
    stack<ll> s;
    ll number = 0;
    s.push(i);
    while (!s.empty())
    {
        ++number;
        ll p = s.top();
        s.pop();
        rep(i, 0, g[p].size())
        {
            ll tmp = g[p][i];
            if (v[tmp] == false)
            {
                v[tmp] = true;
                s.push(tmp);
            }
        }
    }
    return number;
}
ll solver(vector<vector<ll>> &g, ll n)
{
    ll ans = 0;
    ll v[n + 1] = {};
    rep(i, 1, n + 1)
    {
        if (v[i] == false)
        {
            ll tmp = dfs(g, v, i);
            if (tmp > ans)
                ans = tmp;
        }
    }
    return ans;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n, m;
    cin >> n >> m;
    ll a, b;
    vector<vector<ll>> g(n + 1, vector<ll>(0));
    rep(i, 0, m)
    {
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    cout << solver(g, n) << endl;
}