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: Time Limit Exceeded)

import math

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

M=N*K

logNK=math.floor(math.log2(N*K))
if logNK==0 :
    logNK=1
f=([[0]*N for i in range(logNK)])

#calc f0(n)
idx=[[] for i in range(200001)]
for i, a in enumerate(A):
    idx[a].append(i)


for i, a in enumerate(A):
    l=idx[a]
    n=l.index(i)
    if n+1 == len(l):
        j = N + l[0]
    else:
        j = l[n+1]
    f[0][i]=j-i+1


for k in range(1,logNK):
    for i in range(N):
        f[k][i]=f[k-1][ i ] + f[ k-1 ][(i + f[k-1][i])%N ]

#step2
n=0
for k in range(logNK-1,0,-1):
    i=n%N
    if n + f[k][i] < M:
        n+=f[k][i]

#step3
tmp=[]
for i in range(n,M):
    imodn=i%N
    if A[imodn] in tmp:
        tmp=tmp[0:tmp.index(A[imodn])]
    else:
        tmp.append(A[imodn])

print(' '.join(map(str,tmp)))