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 rep(i, n) for (int i = 0; i < n; i++)
#define MOD 1000000007
#define MOD2 998244353
#define INF MOD
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

struct UnionFind {
    vector<int> par;
    vector<int> siz;

    UnionFind(int n) : par(n), siz(n, 1) {
        for (int i = 0; i < n; i++) par[i] = i;
    }

    int root (int i) {
        while (par[i] != i) {
            i = par[i] = par[par[i]];
        }
        return i;
    }
    bool marge(int x, int y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool is_same(int x, int y) {
        return root(x) == root(y);
    }

    int size(int i) {
        return siz[root(i)];
    }
};

int main() {
    int n, m;
    cin >> n >> m;
    UnionFind uf(n);
    map<P, bool> mp;
    rep(i,m) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        /*
        if (mp[make_pair(x, y)]) {
            continue;
        }
        mp[make_pair(x, y)] = true;
        */
        uf.marge(x, y);
    }
    vector<int> a = uf.siz;
    
    set<int> st;
    /*
    rep(i,n) {
        st.insert(a[i]);
    }
    */

    int ans = 1;
    rep(i,n) {
        ans = max(ans, a[i]);
    }

    // cout << st.size() << endl;
    cout << ans << endl;
}