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)

// AtCoder :: Beginner Contest 177 ::
//

#include <bits/stdc++.h>

using namespace std;

void dfs(int node, int compId, const vector<vector<int>> &adj, vector<int> &ids, vector<int> &count) {
    ids[node] = compId;
    count[compId]++;
    for (int neighbor : adj[node]) {
        if (ids[neighbor] == 0) {
            dfs(neighbor, compId, adj, ids, count);
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int nodeCount, edgeCount;
    cin >> nodeCount >> edgeCount;
    vector<vector<int>> adj(nodeCount+1, vector<int>());
    for (int i{0}; i < edgeCount; ++i) {
        int a, b;
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    vector<int> ids(nodeCount+1, 0);
    vector<int> count(nodeCount+1, 0);
    int soln{0};
    for (int start{1}; start <= nodeCount; ++start) {
        if (ids[start] == 0) {
            dfs(start, start, adj, ids, count);
        }
        soln = max(soln, count[start]);
    }
    cout << soln << "\n";
    return EXIT_SUCCESS;
}