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>
typedef long long LL;
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;
LL p = 1000000007ll;

int main(void)
{
  LL n,k;
  cin >> n >> k;
  vector<LL> a(n);
  for(LL &x:a) cin >> x;
  vector<LL> first(212345,-1);
  vector<LL> last(212345,-1);
  vector<LL> succ(n,-1);
  REP(i,n) {
    if(first[a[i]] == -1) first[a[i]] = i;
    if(last[a[i]] != -1) succ[last[a[i]]] = i;
    last[a[i]]=i;
  }

  vector<LL> answers;
  LL tmpans = n;
  do {
    answers.push_back(tmpans);
    LL pos = (tmpans == n) ? 0 : first[a[tmpans]]+1;
    while(true) {
      if(pos==n) break;
      if(succ[pos]==-1) break;
      else pos = succ[pos]+1;
    }
    tmpans = pos;
  } while(tmpans != n);
  LL from = answers[k%answers.size()];
  vector<LL> res;
  set<LL> nums;
  FOR(i,from,n) {
    if(nums.count(a[i])) {
      while(res.back() != a[i]) {
        nums.erase(res.back());
        res.pop_back();
      }
      nums.erase(res.back());
      res.pop_back();
    } else {
      res.push_back(a[i]);
      nums.insert(a[i]);
    }
  }
  REP(i, res.size()) {
    if(i) cout << " ";
    cout << res[i];
  }
  cout << endl;
}
