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)

class UnionFind:
    def __init__(self, N):
        self.N = N
        self.parent = [-1] * N

    def root(self, A):
        if (self.parent[A] < 0):
            return A
        self.parent[A] = self.root(self.parent[A])
        return self.parent[A]

    def size(self, A):
        return -self.parent[self.root(A)]

    def unite(self, A, B):
        A = self.root(A)
        B = self.root(B)

        if (A == B):
            return False

        if (self.size(A) < self.size(B)):
            A, B = B, A

        self.parent[A] += self.parent[B]
        self.parent[B] = A

        return True

    def is_in_same(self, A, B):
        return self.root(A) == self.root(B)


N, M = map(int, input().split())
uf = UnionFind(N)

for i in range(M):
    a, b = map(int, input().split())
    uf.union(a, b)

print(max(uf.size(i)) for i in range(1, N+1))