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.

Examples

Input

3 2
1 2 3


Output

2 3


Input

5 10
1 2 3 2 3


Output

3


Input

6 1000000000000
1 1 2 2 3 3


Output




Input

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


Output

9 2 6
#include <bits/stdc++.h>
using namespace std;

int main() {
  int n;
  long long k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int si = 1;
  while ((1ll << si) <= k * n) si++;
  vector<vector<long long>> par(si, vector<long long>(n, 1e18));
  map<int, int> ma;
  for (int i = n - 1; i >= 0; i--) {
    if (ma.count(a[i])) {
      par[0][i] = ma[a[i]] - i;
    }
    ma[a[i]] = i + 1;
  }
  for (int i = n - 1; i >= 0; i--) {
    if (par[0][i] == 1e18) {
      par[0][i] = ma[a[i]] + n - i;
    }
  }
  for (int i = 0; i < si - 1; i++) {
    for (int j = 0; j < n; j++) {
      if (par[i][j] == 1e18) continue;
      int p = (par[i][j] + j) % n;
      if (par[i][j] + par[i][p] <= k * n) {
        par[i + 1][j] = par[i][j] + par[i][p];
      }
    }
  }
  long long sum = 0;
  for (int i = si - 1; i >= 0; i--) {
    if (sum + par[i][sum % n] <= k * n) {
      sum += par[i][sum % n];
    }
  }
  vector<int> ans;
  while (sum < k * n) {
    if (sum + par[0][sum % n] <= k * n) {
      sum += par[0][sum % n];
    } else {
      ans.push_back(a[sum % n]);
      sum++;
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i];
    cout << (i == ans.size() - 1 ? '\n' : ' ');
  }
  return 0;
}
