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
from collections import deque


def solve(i):
    if i in memo:
        return memo[i]
    q = deque([i])
    while True:
        nxt_i = next_ind[i]
        if nxt_i >= n:
            break
        q.append(nxt_i)
        i = nxt_i
    while q:
        memo[q.pop()] = nxt_i % n
    return memo[i]



n, k = map(int, input().split())
a = list(map(int, input().split())) + [-1, -1]
k_time = k
n += 2
aa = a + a
next_ind = [-1] * (2 * n)
tmp_memo = {}
for i in range(2 * n)[::-1]:
    if aa[i] not in tmp_memo:
        tmp_memo[aa[i]] = i + 1
    else:
        next_ind[i] = tmp_memo[aa[i]]
        tmp_memo[aa[i]] = i + 1

log_size = k.bit_length() + 2
double = [[-1] * n for i in range(log_size)]

# ダブリングで2^0回(= 1回)たどって到達する値を求める
memo = {}
for i in range(n):
    double[0][i] = solve(i)

# ダブリングで2^k回たどって到達する値を求める
for k in range(1, log_size):
    for i in range(n):
        double[k][i] = double[k - 1][double[k - 1][i]]

pos = 0
k_time -= 1
for k in range(log_size):
    if (k_time >> k) & 1:
        pos = double[k][pos]
q = deque([])
set_ = set()
for i in range(pos, n):
    if a[i] not in set_:
        set_.add(a[i])
        q.append(a[i])
    else:
        while True:
            tmp = q.pop()
            set_.remove(tmp)
            if tmp == a[i]:
                break
print(*q)