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 Python (3.8.2) (judge verdict: Wrong Answer)

N, M = map(int, input().split())
relationship = []
reldict = {}
counter = [0]*(N+1)
refer = [0]*(N+1)
for _ in range(M):
    a, b = map(int, input().split())
    if str(a) not in reldict:
        if str(b) not in reldict:
            reldict[str(a)] = 1
            reldict[str(b)] = 2
            refer[a] = a
            refer[b] = a
            counter[a] = 2

        elif str(b) in reldict:
            counter[refer[b]] += 1
            reldict[a] = counter[refer[b]]
            refer[a] = refer[b]

    elif str(a) in reldict:
        if str(b) not in reldict:
            counter[refer[a]] += 1
            reldict[b] = counter[refer[a]]
            refer[b] = refer[a]


        elif b in reldict:
            pass

#print(reldict)
print(max(counter))