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: Accepted)

from collections import Counter

n, k = [int(item) for item in input().split()]
a = [int(item) for item in input().split()]
OVER_RUN = n*k

cnt = [-1] * (2 * 10**5 + 10)
transition = [[-1] * n for _ in range(58)]
transition[0] = [0 for _ in range(1,n+1)]
for i, item in enumerate(a + a):
    if cnt[item] == -1:
        cnt[item] = i
        continue
    if cnt[item] < n:
        transition[1][cnt[item]] = i+1 - cnt[item]
    cnt[item] = i

for i in range(2,58):
    for j in range(n): 
        walk = transition[i-1][j] + transition[i-1][(transition[i-1][j] + j) % n]
        if walk > OVER_RUN:
            transition[i][j] = OVER_RUN
        else:
            transition[i][j] = walk

step = 0
step_mod = 0
while True:
    itr = 57 
    while transition[itr][step_mod] + step >= OVER_RUN:
        itr -= 1
    step += transition[itr][step_mod]
    step_mod = step % n
    if itr == 0:
        break

# Simulation
target = a[step_mod:]
ans = []
cnt = Counter()
for item in target:
    if cnt[item] > 0:
        while True:
            p = ans.pop()
            cnt[p] -= 1
            if p == item or not p:
                break
    else:
        cnt[item] += 1
        ans.append(item)
print(" ".join([str(item) for item in ans]))