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: Wrong Answer)

from collections import deque
N, M = map(int,input().split())
par = [i for i in range(N+1)]
rank = [0] * (N+1)

def find(x):
    if par[x] == x:
        return x
    else:
        # 走査していく過程で親を書き換える
        par[x] = find(par[x])
        return par[x]
#x = find(1)
 # 併合
def union(x, y):
    x = find(x)
    y = find(y)
    if rank[x] < rank[y]:
        par[x] = y
    else:
        par[y] = x
        if rank[x] == rank[y]:
            rank[x] += 1
    # 同じ集合に属するか判定
def same_check(x, y):
    return find(x) == find(y)
    
for m in range(M):
    a, b = map(int, input().split())
    union(a,b)
#print(par)
#or n in range(1,N+1):
mymax = 0
rank = [0] * (N+1)
for n in range(N+1):
    rank[par[n]] += 1
print(max(rank))