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 Lua (Lua 5.3.5) (judge verdict: Accepted)

local n, m = io.read("*n", "*n")
local parent = {}
for i = 1, n do
  parent[i] = i
end

local function uf_findroot(idx)
  local idx_update = idx
  while parent[idx] ~= idx do
    idx = parent[idx]
  end
  while parent[idx_update] ~= idx do
    parent[idx_update], idx_update = idx, parent[idx_update]
  end
  return idx
end

for i = 1, m do
  local a, b = io.read("*n", "*n")
  local ra, rb = uf_findroot(a), uf_findroot(b)
  parent[rb], parent[b] = ra, ra
end
local map = {}
for i = 1, n do
  local r = uf_findroot(i)
  if not map[r] then map[r] = 1
  else map[r] = map[r] + 1
  end
end
local max = 0
for _k, v in pairs(map) do
  max = max < v and v or max
end
print(max)