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
# -*- coding: utf-8 -*-

def solve():
    N, K = map(int, input().split())
    A = list(map(int, input().split()))
    D = {}
    for j,a in enumerate(A):
        D[a] = D.get(a,[])+[j]

    t,k = 0,0
    T = [0]
    while True:
        k = (D[A[t]].index(t)+1)%len(D[A[t]])
        if k == 0:
            T.append(A[t])
        t = (D[A[t]][k]+1)%N
        if t == 0:
            break

    if T == [0]:
        return ''
    if K == 1 or (K-1)%len(T) == 0:
        t = 0
    else:
        t = D[T[(K-1)%len(T)]][0]+1
    S = []
    flag = False
    while t < N:
        k = (D[A[t]].index(t)+1)%len(D[A[t]])
        if k == 0:
            S.append(A[t])
            t += 1
        else:
            t = D[A[t]][k]+1

    res = ' '.join(map(str,S))

    return str(res)

if __name__ == '__main__':
    print(solve())