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
import sys
from collections import defaultdict
from bisect import bisect_right

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

# Aの中で次に同じ数字が出てくるindex
N, K = lr()
A = lr()
next_index = [-1] * N
dict = defaultdict(list)
for i in range(N):
    dict[A[i]].append(i)

for i in range(N):
    a = A[i]
    length = len(dict[a])
    j = bisect_right(dict[a], i)
    next_index[i] = dict[a][j%length]

cur = 0
time = 1
while cur != N:
    ne = next_index[cur]
    if ne <= cur:
        time += 1
    cur = ne + 1
# timeで一周する
K %= time
time = 0
cur = 0
while time < K:
    ne = next_index[cur]
    if ne <= cur:
        time += 1
    if time == K:
        break
    cur = ne + 1

answer = []
while cur < N:
    ne = next_index[cur]
    if ne <= cur:
        answer.append(A[cur])
        cur += 1
    else:
        cur = ne + 1

print(*answer)
# 14