Problem:
Score : 700 points

Problem Statement

We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}).
Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.

Snuke has an integer sequence s, which is initially empty.
For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:

If s does not contain X_i: add X_i to the end of s.

If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.

Find the elements of s after Snuke finished the operations.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq 10^{12}

1 \leq A_i \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 \cdots A_{N-1}

Output

Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.

Sample Input 1

3 2
1 2 3

Sample Output 1

2 3

In this case, X=(1,2,3,1,2,3).
We will perform the operations as follows:

i=0: s does not contain 1, so we add 1 to the end of s, resulting in s=(1).

i=1: s does not contain 2, so we add 2 to the end of s, resulting in s=(1,2).

i=2: s does not contain 3, so we add 3 to the end of s, resulting in s=(1,2,3).

i=3: s does contain 1, so we repeatedly delete the element at the end of s as long as s contains 1, which causes the following changes to s: (1,2,3)→(1,2)→(1)→().

i=4: s does not contain 2, so we add 2 to the end of s, resulting in s=(2).

i=5: s does not contain 3, so we add 3 to the end of s, resulting in s=(2,3).

Sample Input 2

5 10
1 2 3 2 3

Sample Output 2

3

Sample Input 3

6 1000000000000
1 1 2 2 3 3

Sample Output 3

s may be empty in the end.

Sample Input 4

11 97
3 1 4 1 5 9 2 6 5 3 5

Sample Output 4

9 2 6

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

N, K = gets.split.map(&:to_i)
A = gets.split.map(&:to_i)

indices = Hash.new { |h, k| h[k] = [] }

A.each_with_index do |a, idx|
  indices[a] << idx
end

edges = Hash.new

A.uniq.each do |a|
  n = indices[a].size

  n.times do |i|
    idx = indices[a][i]

    edges[idx] = indices[a][(i + 1) % n]
  end
end

head = -1
node = [0, :off]
path = []
visited = Hash.new(false)

# p indices

while !visited[node]
  visited[node] = true
  cur, status = node

  loop_cnt = 0

  loop {
    a = A[cur]

    if loop_cnt == 0 && status == :on
      to = indices[a].first
    else
      to = indices[a].bsearch { |x| x > cur }
    end

    # p [:a, a, :cur, cur, :to, to, :status, status]

    if to.nil?
      case status
      when :on
        status = :off

        if indices[a].size <= 1
          cur = (cur + 1) % N
        else
          cur = indices[a].first
        end

        break if cur == 0
      when :off
        status = :on
        break
      end
    else
      case status
      when :on
        status = :off
        cur = (cur + 1) % N
        break if cur == 0
      when :off
        status = :off
        cur = (to + 1) % N
        break if cur == 0
      end
    end

    loop_cnt += 1
  }

  node = [cur, status]
  # p [:node, node]
  path << node[0]
end

def simulate(init)
  counter = Hash.new(0)

  init.each do |a|
    counter[a] += 1
  end

  res = init.dup

  A.each do |a|
    if counter[a] > 0
      while counter[a] > 0
        v = res.pop
        counter[v] -= 1
      end
    else
      counter[a] += 1
      res << a
    end
  end

  res
end

# p edges
# p indices
path.pop
pl = path.size
# p [:path, path]

if K == 1
  puts simulate([]).join(' ')
elsif pl == 0
  puts ''
else
  v = ((K - 2) % pl)
  idx = path[v]
  a = A[idx]
  # p [:k, k]
  # p simulate([k])
  res = simulate([a])
  puts res.join(' ')
end