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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ll long long int
const int INF = 2147483647;

using namespace std;

int main()
{
  int n;
  ll k;
  cin >> n >> k;
  vector<int> a(n);
  int second = 0;
  bool flag = true;
  map<int, priority_queue<int, vector<int>, greater<int>>> q;
  REP(i, n)
  {
    cin >> a[i];
    q[a[i]].push(i);
    if (i != 0 && flag && a[i] == a[0])
    {
      flag = false;
      second = i;
    }
  }

  map<int, int> min_index;
  map<int, int> next;
  map<int, int> f;
  REP(i, n)
  {
    if (min_index[a[i]] == 0 && a[i] != a[0])
      min_index[a[i]] = i;
  }

  map<int, int> near_index;
  for (int i = n - 1; i >= 0; i--)
  {
    if (near_index[a[i]] == 0)
    {
      next[i] = a[i];
    }
    else if (near_index[a[i]] == n - 1)
    {
      next[i] = 0;
    }
    else
    {
      next[i] = next[near_index[a[i]] + 1];
    }
    near_index[a[i]] = i;
  }

  f[0] = second == 0 ? a[0] : next[second + 1];
  REP(i, n)
  {
    if (f[a[i]] != 0)
      continue;
    f[a[i]] = next[min_index[a[i]] + 1];
  }

  ll loop = 1;
  int tmp = f[0];
  REP(i, n)
  {
    if (tmp == 0)
      break;
    tmp = f[tmp];
    loop++;
  }

  // 解の構成
  tmp = 0;
  for (int i = 0; i < ((k - 1) % loop); i++)
  {
    tmp = f[tmp];
  }
  int init_index = tmp == 0 ? 0 : min_index[tmp] + 1;

  vector<int> ans;
  for (int i = init_index; i < n; i++)
  {
    bool updated = false;
    while (!q[a[i]].empty())
    {
      int po = q[a[i]].top();
      q[a[i]].pop();
      if (po > i)
      {
        i = po;
        updated = true;
        break;
      }
    }
    if (!updated)
    {
      ans.push_back(a[i]);
    }
  }

  REP(i, ans.size())
  {
    cout << ans[i];
    if (i != ans.size() - 1)
    {
      cout << " ";
    }
  }
  cout << endl;
}