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<bits/stdc++.h>
#define MAXN 200005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int N,a[MAXN];
int tobe[MAXN];
bool has[MAXN];
ll K;
vector<int> v,ans;
int main()
{
    scanf("%d%lld",&N,&K);
    for(int i=1;i<=N;i++) scanf("%d",&a[i]);
    for(int i=1;i<=200000;i++) tobe[i]=i;
    for(int i=1;i<=N;i++) swap(tobe[0],tobe[a[i]]);
    int now=0,tot=0;
    do
    {
        now=tobe[now];
        v.push_back(now);
        tot++;
    }while(now!=0);
    reverse(v.begin(),v.end());
    K=K%(int)v.size();
    int x=v[K];
    if(x==0) return 0;
    memset(has,false,sizeof(has));
    for(int i=N;i>=1;i--)
    {
        if(a[i]==x)
        {
            for(int j=i;j<=N;j++)
            {
                if(has[a[j]])
                {
                    while(ans.back()!=a[j]) has[ans.back()]=false,ans.pop_back();
                    ans.pop_back();
                    has[a[j]]=false;
                    continue;
                }
                ans.push_back(a[j]);
                has[a[j]]=true;
            }
            for(auto x:ans) printf("%d ",x);
            return 0;
        }
    }
}
