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: Accepted)

from collections import deque
n, m = map(int, input().split())
lst = [[] for _ in range(n)]
visited = [0] * n
for _ in range(m):
    a, b = map(int, input().split())    
    a -= 1
    b -= 1
    lst[a].append(b)
    lst[b].append(a)

def dfs(st):
    global visited
    global c
    stack = deque([st])
    visited[st] = 1
    c = 1
    while stack:
        nw = stack.pop()
        if visited[nw] == 0:
            c = 1
        for _ in lst[nw]:
            if visited[_] == 0:
                visited[_] = 1
                c += 1
                stack.append(_)
        
ans = 0
for _ in range(n):
    if visited[_] == 0:
        dfs(_)
        ans = max(ans, c)    
print(ans)