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 unsigned int ui;
typedef pair<int, int> pi;

#define fi first
#define se second

ll read() {
  ll res = 0, w = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') c = getchar(), w = -1;
  while (isdigit(c)) res = (res << 1) + (res << 3) + c - '0', c = getchar();
  return res * w;
}

const int N = 2e5 + 10;

int st[N], p[N], a[N], nxt[N], las[N], x[N], n, pos, top;
ll k;

int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    nxt[las[a[i]]] = i;
    las[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    nxt[las[a[i]]] = i;
    las[a[i]] = i;
  }
  for (int i = 1; i < n; i++) x[i] = i + 1;
  x[n] = 1;
  st[++top] = 1, p[top] = 1, pos = 1;
  do {
    ++top, p[top] = p[top - 1] + (nxt[pos] <= pos) + (nxt[pos] == n);
    st[top] = pos = x[nxt[pos]];
  } while (st[top] != st[1]);
  k -= (k - 1) / (p[top] - 1) * (p[top] - 1);
  int j = 1;
  while (j <= top - 1 && p[j + 1] <= k) ++j;
  for (int i = st[j]; i <= n; ++i) {
    if (nxt[i] > i)
      i = nxt[i];
    else
      printf("%d ", a[i]);
  }
  return 0;
}