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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

ll n, k;
unordered_map<int, int> l;

void mul(vector<int> & a, vector<int> & b1, vector<int> & b2) {
  for (int i = n; i >= 0; i--) {
    if (b1[i] == n)
      b1[i] = b2[0];
    else
      b1[i] = b2[l[a[b1[i]]] + 1];
  }
}

void powF(vector<int> & a, vector<int> & b, vector<int> & r, ll e) {
  vector<int> p(n + 1), pp(n + 1);
  p = b;
  while ((e & 1) == 0) {
    pp = p;
    mul(a, p, pp);
    e /= 2;
  }
  r = p;
  e /= 2;
  while (e) {
    pp = p;
    mul(a, p, pp);
    if (e & 1)
      mul(a, r, p);
    e /= 2;
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++)
    cin >> a[i];
  vector<int> b(n + 1), r(n + 1), c(n + 1);
  b[n] = n, c[n] = n;
  for (int i = n - 1; i >= 0; i--) {
    if (l.find(a[i]) == l.end())
      b[i] = i, c[i] = b[i + 1];
    else
      b[i] = b[l[a[i]] + 1], c[i] = c[b[i]];
    l[a[i]] = i;
  }
  powF(a, b, r, k);
  int i = r[0];
  if (i == n)
    cout << "\n";
  while (i < n) {
    cout << a[i] << (c[i] < n ? " " : "\n");
    i = c[i];
  }
}
