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)

import sys
input = sys.stdin.readline
ins = lambda: input().rstrip()
ini = lambda: int(input().rstrip())
inm = lambda: map(int, input().rstrip().split())
inl = lambda: list(map(int, input().split()))
out = lambda x: print('\n'.join(map(str, x)))

def dfs(x):
    stack = [x]
    visited = 0
    checked[x] = 1
    while stack:
        visited += 1
        node = stack.pop()
        for i in graph[node]:
            if checked[i] == 0:
                stack.append(i)
                checked[i] = 1
    return visited
 
n, m = inm()
checked = [0] * (n+1)
graph = {}
for i in range(m):
    a, b = inm()
    if a not in graph:
        graph[a] = set()
    if b not in graph:
        graph[b] = set()
    graph[a].add(b)
    graph[b].add(a)
ans = 0
for i in graph:
    if checked[i] == 0:
        ans = max(ans, dfs(i))
print(ans)