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;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  int n; lint k; cin >> n >> k;
  V<> a(n); for (auto&& e : a) cin >> e;
  map<int, int> mp;
  V<> nxt(n);
  for (int i = 2 * n - 1; i >= 0; --i) {
    if (i < n) nxt[i] = mp[a[i]];
    mp[a[i % n]] = i;
  }

  V<lint> vis(n, -1);
  V<> res;
  lint p = 0;
  bool first = true;
  while (p < n * k) {
    if (first and vis[p % n] != -1) {
      lint c = (p - vis[p % n]);
      p += (n * k - p) / c * c;
      first = false;
      continue;
    }
    vis[p % n] = p;
    lint q = p / n * n + nxt[p % n] + 1;
    if (q > n * k) {
      res.push_back(a[p++ % n]);
    } else {
      p = q;
    }
  }

  int m = res.size();
  for (int i = 0; i < m; ++i) cout << res[i] << " \n"[!(m + ~i)];
}