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
import java.util.*;

public class Main {

    static final int NONE = -1;
    static final int A_MAX = 2 * 100000;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long K = sc.nextLong();
        int[] A = new int[N];

        for (int i = 0; i < N; i++) A[i] = sc.nextInt();

        int[] toNext = new int[N];
        int[] first = new int[A_MAX + 1];
        int[] last = new int[A_MAX + 1];
        Arrays.fill(first, -1);
        Arrays.fill(last, -1);

        for (int i = 0; i < N; i++) {
            if (first[A[i]] == NONE) {
                first[A[i]] = i;
                last[A[i]] = i;
            } else {
                toNext[last[A[i]]] = i - last[A[i]] + 1;
                last[A[i]] = i;
            }
        }

        for (int i = 0; i < N; i++) {
            if (first[A[i]] != NONE) {
                toNext[last[A[i]]] = N - last[A[i]] + first[A[i]] + 1;
            }
        }

        int pos = 0;
        long ops = 0;
        long[] firstVisit = new long[N];
        Arrays.fill(firstVisit, NONE);

        while (true) {
            if (firstVisit[pos] != NONE) {
                long cycleLength = ops - firstVisit[pos];
                ops += ((N * K - ops) / cycleLength) * cycleLength;
            } else {
                firstVisit[pos] = ops;
            }
            if (ops + toNext[pos] > N * K) break;
            ops += toNext[pos];
            pos = (pos + toNext[pos]) % N;
        }

        long remainingOps = N * K - ops;

        LinkedList<Integer> sol = new LinkedList<>();
        boolean[] contains = new boolean[A_MAX + 1];

        for (int i = 0; i < remainingOps; i++) {
            int a = A[(pos + i) % N];
            if (contains[a]) {
                int x = sol.pollLast();
                while (x != a) {
                    contains[x] = false;
                    x = sol.pollLast();
                }
                contains[x] = false;
            } else {
                contains[a] = true;
                sol.addLast(a);
            }
        }

        for (int s : sol) {
            System.out.print(s + " ");
        }
    }
}
