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;
typedef long long ll;
typedef long double ld;

int a[200000];
const int lods = 100;
ld nxt[lods][200000];
int last[200001];

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  ll n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  fill_n(last, 200001, n);
  for (int i = n-1; i >= 0; i--) {
    last[a[i]] = i+n;
  }
  for (int i = n-1; i >= 0; i--) {
    nxt[0][i] = last[a[i]]-i+1;
    last[a[i]] = i;
  }
  for (int l = 0; l+1 < lods; l++) {
    for (int i = 0; i < n; i++) {
      if (nxt[l][i] > 1e18) {
	nxt[l+1][i] = 1e18;
	continue;
      }
      nxt[l+1][i] = nxt[l][i]+nxt[l][(i+(ll)nxt[l][i])%n];
    }
  }
  ll p = 0;
  for (int l = lods-1; l >= 0; l--) {
    while (p+nxt[l][p%n] < n*k) {
      p += (ll)nxt[l][p%n];
    }
  }
  vector<int> ans;
  set<int> in;
  for (int i = p%n; i < n; i++) {
    if (in.count(a[i])) {
      while (in.count(a[i])) {
	in.erase(ans.back());
	ans.pop_back();
      }
    } else {
      ans.push_back(a[i]);
      in.insert(a[i]);
    }
  }
  for (int i : ans) cout << i << ' ';
  cout << endl;
}
