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: Runtime Error)

class disjoint:
    def __init__(self, x: int):
        self.a = [-1] * x

    def root(self, x: int):
        if self.a[x] < 0:
            return x
        self.a[x] = self.root(self.a[x])
        return self.a[x]

    def unite(self, x: int, y: int):
        x, y = self.root(x), self.root(y)
        if x == y:
            return
        if x < y:
            x, y = y, x
        self.a[x] += self.a[y]
        self.a[y] = x

    def size(self, x: int):
        if self.a[x] < 0:
            return -self.a[x]
        return self.size(self.a[x])


n, m = [int(i) for i in input().split()]
solution = disjoint(n)
for _ in range(m):
    a, b = [int(i) for i in input().split()]
    a -= 1
    b -= 1
    solution.unite(a, b)

ans = 1
for i in range(n):
    ans = max(ans, solution.size(i))

print(int(ans))