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: Runtime Error)

n,t = list(map(int,input().split()))
graph = [[] for _ in range(n)]
visited = [0]*n
for _ in range(t):
    a,b = list(map(int,input().split()))
    a -= 1;b -= 1
    graph[a].append(b)
    graph[b].append(a)
def dfs(now):
    visited[now] = 1
    num = 1
    for i in graph[now]:
        if not visited[i]:
            num += dfs(i)
    return num
ans = 0
for i in range(n):
    if not visited[i]:
        ans = max(ans,dfs(i))

print(ans)

# allsets = []
# n,t = list(map(int,input().split()))
# for _ in range(t):
#     a,b = list(map(int,input().split()))
#     if a < b and [a,b] not in allsets:
#         allsets.append([a,b])
#     elif b < a and [b,a] not in allsets:
#         allsets.append([b,a])
# # print(allsets)
# hmap = {}
# visitedNode = {}
# for a,b in allsets:
#     if a in visitedNode:
#         if b not in hmap[visitedNode[a]]:
#             hmap[visitedNode[a]].append(b)
#         del visitedNode[a]
#         continue
#     if a in hmap:
#         hmap[a].append(b)
#         visitedNode[b] = a
#     else:
#         hmap[a] = [b]
#         visitedNode[b] = a
# # print(hmap)
# res = 0
# for k,v in hmap.items():
#     res = max(res,len(v)+1)
# print(res)