Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3 2
1 2 3

Sample Output 1

2 3

In this case, X=(1,2,3,1,2,3).
We will perform the operations as follows:

i=0: s does not contain 1, so we add 1 to the end of s, resulting in s=(1).

i=1: s does not contain 2, so we add 2 to the end of s, resulting in s=(1,2).

i=2: s does not contain 3, so we add 3 to the end of s, resulting in s=(1,2,3).

i=3: s does contain 1, so we repeatedly delete the element at the end of s as long as s contains 1, which causes the following changes to s: (1,2,3)→(1,2)→(1)→().

i=4: s does not contain 2, so we add 2 to the end of s, resulting in s=(2).

i=5: s does not contain 3, so we add 3 to the end of s, resulting in s=(2,3).

Sample Input 2

5 10
1 2 3 2 3

Sample Output 2

3

Sample Input 3

6 1000000000000
1 1 2 2 3 3

Sample Output 3

s may be empty in the end.

Sample Input 4

11 97
3 1 4 1 5 9 2 6 5 3 5

Sample Output 4

9 2 6

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from bisect import bisect_right as bisect
N, K = map(int, input().split())
A = list(map(int, input().split()))
m = max(A)
a_to_i = [[] for _ in range(m+1)]
for i in range(N):
    a_to_i[A[i]].append(i)
now = 0
next_k_i = [[None]*N for _ in range(60)]
for now in range(N):
    h = 0
    i = bisect(a_to_i[A[now]], now)
    if i == len(a_to_i[A[now]]):
        i = 0
        h+=N
    i = a_to_i[A[now]][i]
    h+=(i-now+1)
    next_k_i[1][now] = ((i+1)%N, h)
for i in range(2, 60):
    for j in range(N):
        a, b = next_k_i[i-1][j]
        c, d = next_k_i[i-1][a]
        next_k_i[i][j] = (c, b+d)
now = 0
cnt = K*N
while cnt:
    flag = False
    for i in range(1, 60)[::-1]:
        a, b = next_k_i[i][now]
        if cnt>=b:
            cnt-=b
            now = a
            flag = True
            break
    if not flag:
        break
ans = []
while now<N:
    a = next_k_i[1][now][0]
    if next_k_i[1][now][0] == 0:
        a+=N
    if a>now+1:
        now = a
        continue
    ans.append(A[now])
    now+=1
print(*ans)