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 Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

N, M = gets.split.map(&:strip).map(&:to_i)

A = []
B = []

M.times do
  a, b = gets.split.map(&:strip).map(&:to_i)
  A << a
  B << b
end

gs = {}
@dg = {}
g = 0
mn = 1
def same(g1, g2)
  g1 == g2 || g1 == @dg[g2] || @dg[g1] == g2
end

M.times do |i|
  a = A[i]
  b = B[i]
  if !gs[a]
    if !gs[b]
      g += 1
      gs[a] = g
      gs[b] = g
    else
      gs[a] = gs[b]
    end
  else
    if !gs[b]
      gs[b] = gs[a]
    else
      if !same(gs[a],gs[b])
        @dg[gs[a]] = gs[b]
        @dg[gs[b]] = gs[a]
      end
    end
  end
end



rg = {}
ng = g
(1..g).each do |g|
  if sg = @dg[g]
    ng += 1
    rg[g] = ng
    while sg
      rg[sg] = ng
      sg = @dg[sg]
    end
  else
    rg[g] = g
  end
end

m = 1
c = {}
gs.each do |_, g|
  gg = rg[g]
  c[gg] ||= 0
  c[gg] += 1
  if c[gg] > m
    m = c[gg]
  end
end

puts m