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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

N, M = map(int, input().split())
lst = [0] * N
num = 1
for i in range(M):
    A, B = map(int, input().split())
    A -= 1
    B -= 1
    if lst[A] != 0 and lst[B] != 0:     # 集合の結合(完全グラフ化)
        if lst[A] != lst[B]:
            for j in range(N):
                if lst[j] == lst[B]:
                    lst[j] = lst[A]
    elif lst[A] == 0 and lst[B] == 0:
        lst[A] = num
        lst[B] = num
        num += 1
    else:
        if lst[A] == 0:
            lst[A] = lst[B]
        else:
            lst[B] = lst[A]
    # print(A, B, lst)

cnt_lst = [0] * num
for i, value in enumerate(lst):
    cnt_lst[value] += 1
# print(cnt_lst)
ans = 0
for i, value in enumerate(cnt_lst):
    if i != 0: ans = max(ans, value)
if num == 1:
    ans = 1
print(ans)