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.par=[i for i in range(N)]
        self.siz=[1 for _ in range(N)]
        self.rank=[0 for _ in range(N)]
    def find(self,x):
        if self.par[x]==x:
            return x
        else:
            self.par[x]=self.find(self.par[x])
            return self.par[x]
    def union(self,a,b):
        a=self.find(a)
        b=self.find(b)
        if a==b:
            return 0
        else:
            if self.rank[a]>self.rank[b]:
                self.par[b]=a
                self.siz[a]+=self.siz[b]
            else:
                self.par[a]=b
                self.siz[b]+=self.siz[a]
                if self.rank[a]==self.rank[b]:
                    self.rank[b]+=1
    def size(self,a):
        return self.siz[self.find(a)]
    def same(self,a,b):
        return self.find(a)==self.find(b)

N,M=map(int,input().split())
UF=UnionFind(N)
for i in range(M):
    A,B=map(lambda x:int(x)-1,input().split())
    UF.union(A,B)
ans=1
for i in range(N):
    ans=max(ans,UF.size(i))
print(ans)