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: Wrong Answer)

def solve(array):
    ans = []
    memo1 = {}
    for i in range(len(array)):
        if array[i] not in memo1:
            memo1[array[i]] = 1
        else:
            memo1[array[i]] += 1
    i = 0
    while i < len(array):
        if memo1[array[i]] == 1:
            ans.append(array[i])
            i += 1
        else:
            tmp = array[i]
            memo1[array[i]] -= 1
            i += 1
            while True:
                memo1[array[i]] -= 1
                if tmp == array[i]:
                    i += 1
                    break
                else:
                    i += 1
    return ans   

n, k = map(int, input().split())
a = list(map(int, input().split()))

start = {}
for i in range(len(a))[::-1]:
    start[a[i]] = i
flag = False
li = []
tmp = []
while True:
    memo = {}
    if not tmp:
        start_num = 0
    else:
        start_num = start[tmp[-1]] + 1
    tmp = solve(a[start_num:n])
    #print(tmp)
    if li and li[-1] == tmp:
        flag = True
        break
    li.append(tmp)
    if not tmp:
        break

t = 1
for i in range(len(li)):
    t += len(li[i])
if flag:
    t -= 1
#print(li)
#print(t)
ans = k % t
if ans == 0:
    print()
else:
    for i in range(len(li)):
        if ans <= len(li[i]):
            print(" ".join(map(str, li[i][ans-1:])))
            break
        ans -= len(li[i])