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 Python (3.4.3) (judge verdict: Wrong Answer)

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

if N==1:
    print(A[0] if K%2 else '')
    exit()

AA = A+A
last_i = {}
nx = [-1]*N
for i,a in enumerate(reversed(AA)):
    i = 2*N-i-1
    if i < N:
        nx[i] = (last_i[a] - i) % N + 1
    last_i[a] = i

visited = [-1]*N
visited[0] = 0
j = i = 0
while 1:
    p = nx[i]+N if nx[i]==1 else nx[i]
    j += p
    if visited[j%N] < 0:
        visited[j%N] = j
    else:
        break
    i = j%N
cycle = (j - visited[j%N]) // N

K %= cycle
j = i = 0
while 1:
    p = nx[i]+N if nx[i]==1 else nx[i]
    if j+p > N*K:
        break
    j += p
    i = j%N

if j%N == 0:
    print()
else:
    ans = []
    anss = set()
    for a in A[j%N:]:
        if a in anss:
            while ans[-1] != a:
                anss.remove(ans[-1])
                ans.pop()
            anss.remove(a)
            ans.pop()
        else:
            anss.add(a)
            ans.append(a)
    print(*ans)