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)

n, m = map(int, input().split())

root = [-1]*n#各数字の親を-1で初期化
def r(x):
    if root[x] < 0: #親がなければ自分自身の値を返す
        return x
    else:
        root[x] = r(root[x])#親を辿る
        return root[x]

def unite(x, y):
    x = r(x)
    y = r(y)
    if x == y:#親が同じ場合
        return
    root[x] += root[y]
    root[y] = x #xをyの親とする

for i in range(m):
    a, b = map(int, input().split())
    a -= 1#0_indexにする
    b -= 1
    unite(a, b)

ans = 0
for i in range(n):
    ans = max(ans, -root[i])#要素数が最大の親が答え

print(ans)