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 defaultdict

N, K = map(int, input().split())
A = list(map(int, input().split()))

D = defaultdict(list)

for i in range(N):
    D[A[i]].append(i)

next = [None for _ in range(N)]
dist = [None for _ in range(N)]

for v in D.values():
    l = len(v)
    for i in range(l):
        next[v[i]] = (v[(i + 1) % l] + 1) % N
        dist[v[i]] = (v[(i + 1) % l] - v[i] + N - 1) % N + 2

period = 0
node = 0

while node or not period:
    period += dist[node]
    node = next[node]

rem = N * K % period
count = 0
node = 0

while count + dist[node] <= rem:
    count += dist[node]
    node = next[node]

s = []
used = [0 for _ in range(10**6)]

for i in range(rem - count):
    j = (node + i) % N
    if not used[A[j] - 1]:
        used[A[j] - 1] = 1
        s.append(A[j])
    else:
        while True:
            p = s.pop()
            used[p - 1] = 0
            if p == A[j]:
                break

print(*s)