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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const ll INF = 1e18;

int main() {
  ll n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    a.push_back(a[i]);
  }

  ll nk = n * k;
  int lognk = 0;
  for (ll t = nk; t > 1; t >>= 1) lognk++;
  
  map<int, int> nxt;
  vector<vector<ll>> dbl(lognk + 1, vector<ll>(n, INF));
  for (int i = 2 * n - 1; i >= 0; i--) {
    if (i < n) dbl[0][i] = nxt[a[i]] - i + 1;
    nxt[a[i]] = i;
  }

  for (int k = 0; k < lognk; k++) {
    for (int i = 0; i < n; i++) {
      if (dbl[k][i] == INF) continue;
      dbl[k + 1][i] = dbl[k][i] + dbl[k][(i + dbl[k][i]) % n];
      if (dbl[k + 1][i] > INF) dbl[k + 1][i] = INF;
    }
  }
  
  ll p = 0;
  for (int k = lognk - 1; k >= 0; k--) {
    if (p + dbl[k][p % n] > nk) continue;
    p += dbl[k][p % n];
  }

  vector<int> ans;
  set<int> st;
  for (int i = p; i < nk; ++i) {
    int x = a[i % n];
    if (st.find(x) != end(st)) {
      while (st.find(x) != end(st)) {
        st.erase(ans.back());
        ans.pop_back();
      }
    } else {
      ans.push_back(x);
      st.insert(x);
    }
  }

  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << ((i + 1 == ans.size()) ? '\n' : ' ');
  }
}