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 <cstring>
#include <vector>
using namespace std;
const int N = 200000 + 10;
typedef long long LL;

int n,a[N],nex[N][60],pos[N],lef[N]; LL k;

int main() {
    scanf("%d%lld",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=n;i>=1;i--){
        lef[a[i]]=i;
    }
    memset(pos,0,sizeof(pos));
    nex[n+1][0]=1;
    for(int i=n;i>=1;i--){
        if(!pos[a[i]]) {
            nex[i][0]=lef[a[i]]+1;
        } else {
            nex[i][0]=nex[pos[a[i]]+1][0];
        }
        pos[a[i]]=i;
        //printf("nex[%d]=%d\n", i,nex[i][0]);
    }
    
    for(int i=1;i<60;i++){
        for(int j=1;j<=n+1;j++)nex[j][i]=nex[nex[j][i-1]][i-1];
    }
    int p=1;
    for(int i=59;i>=0;i--){
        if((k-1)>>i&1) {
            p=nex[p][i];
        }
    }
    //printf("p=%d\n", p);
    int stk[N]={0},top=0,has[N]={0};
    for(int i=p;i<=n;i++){
        if(has[a[i]]==0){
            stk[++top]=a[i]; has[a[i]]=1;
        } else {
            while(top) {
                int x=stk[top];
                has[stk[top]]=0; --top;
                if(x==a[i]) break;
            }
        }
    }
    for(int i=1;i<=top;i++) printf("%d ", stk[i]);
}