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 <stdio.h>
#include <set>
#include <utility>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, now = 0, p = 0;
  scanf("%lld%lld", &n, &k);
  --k;
  ll a[n], nex[n], f[n];
  bool flg = false;
  set< pair<ll, ll> > s;
  set< pair<ll, ll> >::iterator it;
  pair<ll, ll> t;
  for(i = 0; i < n; ++i) f[i] = -1;
  for(i = 0; i < 2 * n; ++i) {
    if(i < n) scanf("%lld", &a[i]);
    t = make_pair(a[i % n], 0);
    it = s.lower_bound(t);
    if(it != s.end() && (*it).first == a[i % n]) {
      t = *it;
      s.erase(it);
      if(t.second < n) nex[t.second] = i;
    }
    t.second = i;
    s.insert(t);
  }
  f[0] = 0;
  while(now < k) {
    p = nex[p] + 1;
    if(p >= n) now++, p -= n;
    if(!p) break;
    if(p < n) f[p] = now;
    else {
      ++now, p = 0;
      flg = true;
      break;
    }
  }
  if(!p && k) k %= now;
  if(flg && k == now - 1) return 0;
  for(i = 0; i < n; ++i) if(k == f[i]) break;
  for(; i < n; ++i) {
    if(nex[i] < n) i = nex[i];
    else printf("%lld ", a[i]);
  }
  return 0;
}