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>
#define watch(x) cout << (#x) << " is " << (x) << endl
#define endl "\n"
typedef long long ll;
using namespace std;
int static fast = [](){
  ios::sync_with_stdio(false);
  cin.tie(0); cout.tie(0); return 0;
}();
// freopen("input.txt", "r", stdin);

int find(vector<int>& grp, int x) {
    if (grp[x] == x)
        return x;
    return grp[x] = find(grp, grp[x]);
}

int main() {
    int n, m, u, v;
    cin >> n >> m;
    vector<int> grp(n, 0);
    for(int i = 0; i < n; i++)
        grp[i] = i;
    for(int i = 0; i < m; i++) {
        cin >> u >> v;
        int r1 = find(grp, u-1);
        int r2 = find(grp, v-1);
        grp[r1] = grp[r2] = min(r1, r2);
    }
    int ans = 0;
    unordered_map<int, int> member;
    for(int i = 0; i < n; i++) {
        int r = find(grp, i);
        member[r] += 1;
        ans = max(ans, member[r]);
    }
    cout << ans << endl;
    return 0;
}