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
#include<iostream>
#include<queue>

using namespace std;

long long Del[60][200005],Lastpoz[200005],Ap[200005];
int A[200005];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    long long n,k;
    cin>>n>>k;
    for(int i=1; i<=n; i++)
        cin>>A[i];

    for(int i=n; i>=1; i--)
        Lastpoz[A[i]]=n+i;
    for(int i=n; i>=1; i--)
    {
        Del[0][i]=Lastpoz[A[i]]-i+1;
        Lastpoz[A[i]]=i;
    }

    long long l=n*k;
    for(int i=1; i<=57; i++)
        for(int j=1; j<=n; j++)
            Del[i][j]=min(Del[i-1][j]+Del[i-1][(j+Del[i-1][j]-1)%n+1],l);

    long long poz=1;
    for(int i=57; i>=0; i--)
    {
        if(poz+Del[i][(poz-1)%n+1]<=l)
            poz+=Del[i][(poz-1)%n+1];
    }

    deque<int> Q;
    while(poz<=l)
    {
        if(Ap[A[(poz-1)%n+1]])
        {
            while(!Q.empty() && Ap[A[(poz-1)%n+1]])
            {
                Ap[Q.back()]=0;
                Q.pop_back();
            }
        }
        else
        {
            Ap[A[(poz-1)%n+1]]=1;
            Q.push_back(A[(poz-1)%n+1]);
        }
        poz++;
    }

    while(!Q.empty())
    {
        cout<<Q.front()<<" ";
        Q.pop_front();
    }
    cout<<"\n";
    return 0;
}
