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)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * (n)
        
    def find(self, x):
        if self.parents[x] < 0:     # 根ならその番号を返す
            return x
        else:                       # 根でないならその要素で再検索
            self.parents[x] = self.find(self.parents[x])    # 走査過程で親を書き換え
            return self.parents[x]
        
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:                  # 既に同じ木に属している場合
            return
        
        # 違う木に属している場合は高い方の木に低い方を併合
        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]  # 併合される側の要素数を加算
        self.parents[y] = x                 # 併合する側の親を書き換え
    
    def size(self, x):
        return -self.parents[self.find(x)]
    
    def issame(self, x, y):
        return self.find[x] == self.find(y)


'''
参考
https://qiita.com/K_SIO/items/03ff1fc1184cb39674aa#d-friends
'''
if __name__ == '__main__':
    N, M = map(int, input().split())
    uf = UnionFind(N)

    for _ in range(M):
        A, B = map(int, input().split())
        uf.union(A-1, B-1)
    
    ans = 0

    for i in range(N):
        ans = max(ans, uf.size(i))
    
    print(ans)