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>
using namespace std;
typedef long long ll;
const int N = 2e5+5;
ll n,k;
ll a[200005];
ll q[1000005],ind[N],r=0,f[N][50],pre[N],nxt[N];
//倍增跳的行数
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>k;k--;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        if(!pre[a[i]])pre[a[i]]=i;
    }
    f[n+1][0]=1;
    for(int i=n;i>=1;i--){
        if(!nxt[a[i]]){
            f[i][0]=pre[a[i]]+1;
        }else{
            f[i][0]=f[nxt[a[i]]+1][0];
        }
        nxt[a[i]]=i;
    }
    for(int j=1;j<=40;j++){
        for(int i=1;i<=n+1;i++){
            f[i][j]=f[f[i][j-1]][j-1];
        }
    }
    int now=1;
    for(int i=40;i>=0;i--){
        if((1ll<<i)<=k)k-=1ll<<i,now=f[now][i];
    }
    for(int i=now;i<=n;i++){
        if(ind[a[i]]){
            int p=ind[a[i]];
            for(int j=ind[a[i]];j<=r;j++)
                ind[q[j]]=0;
            r=p-1;
        }else{
            q[++r]=a[i];
            ind[a[i]]=r;
        }
    }
    for(int i=1;i<=r;i++)cout<<q[i]<<' ';
}