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 PyPy3 (7.3.0) (judge verdict: Accepted)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [i for i in range(n)]
        self.size = [1 for _ in range(n)]

    def find(self, x):
        root = x
        while self.parents[root] != root:
            root = self.parents[root]
        return root

    def unite(self, x, y):
        xroot = self.find(x)
        yroot = self.find(y)
        if xroot == yroot: return
        if self.size[xroot] <= self.size[yroot]:
            self.parents[xroot] = yroot
            self.size[yroot] += self.size[xroot]
        else:
            self.parents[yroot] = xroot
            self.size[xroot] += self.size[yroot]

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

uf = UnionFind(N)

for i in range(M):
    a, b = map(int, input().split())
    uf.unite(a - 1, b - 1)

print(max(uf.size))