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.

Examples

Input

3 2
1 2 3


Output

2 3


Input

5 10
1 2 3 2 3


Output

3


Input

6 1000000000000
1 1 2 2 3 3


Output




Input

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


Output

9 2 6
import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda:sys.stdin.readline().rstrip()
def resolve():
    n,k=map(int,input().split())
    A=list(map(int,input().split()))
    idx=[[] for _ in range(max(A)+1)]
    for i,a in enumerate(A):
        idx[a].append(i)

    # いくつ先に移動するかを求める
    next=[None]*n
    for I in idx:
        if(not I): continue
        s=len(I)
        I.append(I[0]+n)
        for i in range(s):
            next[I[i]]=I[i+1]-I[i]+1

    # doubling
    M=10**18 # n*k より大きな数で抑えておく
    nexts=[next]
    for _ in range(60):
        next=nexts[-1]
        nnext=[min(M,next[i]+next[(i+next[i])%n]) for i in range(n)]
        nexts.append(nnext)

    now=0
    for next in nexts[::-1]:
        if(now+next[now%n]<=n*k):
            now+=next[now%n]
    if(now==n*k):
        return

    # A[now%n:] に対してシミュレーションする
    ans=[]
    have=[0]*(max(A)+1)
    for a in A[now%n:]:
        if(not have[a]):
            ans.append(a)
            have[a]=1
        else:
            while(1):
                x=ans.pop()
                have[x]=0
                if(x==a): break

    print(*ans)
resolve()