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++ (Clang 10.0.0) (judge verdict: Runtime Error)

#include <iostream>
#include <set>
#include <vector>
#include <map>
using namespace std;

int main() {
  int64_t n, k, start = 0;
  cin >> n >> k;
  int a[2 * n], next[n];
  for (int i = 0; cin >> a[i]; ++i) {
    a[i + n] = a[i];
  }
  map<int, int> index;
  if (k > 1) {
    for (int i = 2 * n - 1; i >= n; --i) {
      index[a[i]] = i;
    }
  } else {
    for (int i = 2 * n - 1; i >= n; --i) {
      index[a[i]] = n;
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    auto it = index.find(a[i]);
    next[i] = it->second - i + 1;
    it->second = i;
  }
  do {
    if (start + next[start % n] < n * k) {
      start += next[start % n];
    } else {
      break;
    }
  } while (start % n != 0);
  set<int> set;
  vector<int> result;
  for (auto i = n * k / start * start; i < n * k; ++i) {
    auto it = set.find(a[i % n]);
    if (it == end(set)) {
      set.emplace(a[i % n]);
      result.emplace_back(a[i % n]);
    } else {
      while (result.back() != *it) {
        set.erase(result.back());
        result.pop_back();
      }
      result.pop_back();
      set.erase(it);
    }
  }
  for (auto i : result) {
    cout << i << ' ';
  }
  return 0;
}