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)

N, M = map(int,input().split())
AB = [list(map(int,input().split())) for x in range(M)]

parents = [-1]*(N+1)

def find(x):
    global parents
    if parents[x] < 0:
        return x
    else:
        return find(parents[x])

def union(x,y):#xにyを合体させる
    global parents
    X = find(x)
    Y = find(y)
    if X == Y:
        return
    if X > Y:
        X,Y = Y,X

    parents[X] += parents[Y]
    parents[Y] = X



for p in range(M):
    union(AB[p][0],AB[p][1])



print(min(parents)*-1)














"""
F = [[]]
L = []

for x in range(N+1):
    F += [[x+1]]

for p in range(M):
    F[AB[p][0]] += [AB[p][1]]
    F[AB[p][1]] += F[AB[p][0]]

for q in range(N+1):
    F[q] = list(set(F[q]))

for r in range(N+1):
    L += [len(F[r])]

print(L)
print(max(L))
"""