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=0):
        self.d = [-1]*n
        self.u = n
    def root(self, x):
        if self.d[x] < 0:
            return x
        self.d[x] = self.root(self.d[x])
        return self.d[x]
    def unite(self, x, y):
        x, y = self.root(x), self.root(y)
        if x == y:
            return False
        if x > y:
            x, y = y, x
        self.d[x] += self.d[y]
        self.d[y] = x
        self.u -= 1
        return True
    def same(self, x, y):
        return self.root(x) == self.root(y)
    def size(self, x):
        return -self.d[self.root(x)]
    def num_union(self):
        return self.u


import sys
input = sys.stdin.buffer.readline
N, M = map(int, input().split())
uf = UnionFind(N)
for _ in [0]*M:
    a, b = map(int, input().split())
    a -= 1; b -= 1
    uf.unite(a, b)

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