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)

from collections import deque
N,M = map(int,input().split())
graph = [[] for _ in range(N)]
for i in range(M):
    x,y = map(int,input().split())
    x -= 1
    y -= 1
    graph[x].append(y)
    graph[y].append(x)

seen = [None]*N
ans = -1

for i in range(N):
    if seen[i]:
        continue
    seen[i] = True
    tmp = 0
    d = deque()
    d.append(i)
    while d:
        v = d.popleft()
        for i in graph[v]:
            if seen[i]:
                continue
            seen[i] = True
            d.append(i)
            tmp += 1
    ans = max(ans,tmp)
print(ans+1)