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: Accepted)

# import sys
# input = sys.stdin.readline
import collections

def main():
    n, m = input_list()
    Friend = [[] for i in range(n+1)]
    for _ in range(m):
        a, b = input_list()
        Friend[a].append(b)
        Friend[b].append(a)
    Flag = [False for i in range(n+1)]
    ans = 1
    for i in range(1, n+1):
        if Flag[i]:
            continue
        q = collections.deque()
        q.append(i)
        cnt = 1
        while q:
            st = q.popleft()
            Flag[st] = True
            for j in Friend[st]:
                if Flag[j]:
                    continue
                cnt += 1
                Flag[j] = True
                q.append(j)
            ans = max(ans, cnt)
    print(ans)


def input_list():
    return list(map(int, input().split()))


def input_list_str():
    return list(map(str, input().split()))


if __name__ == "__main__":
    main()