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 For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define pii pair<int,int>
#define fi first
#define se second
#define PB push_back
#define ll long long
using namespace std;
const int N=400005;
int a[N],las[N],q[N],n;
ll b[N][60],k;
int main(){
	scanf("%d%lld",&n,&k);
	For(i,1,n) scanf("%d",&a[i]),a[i+n]=a[i];
	Rep(i,n+n,n+1) las[a[i]]=i;
	Rep(i,n,1){
		b[i][0]=las[a[i]];
		las[a[i]]=i;
	}
	For(i,1,58) For(j,1,n)
		if (b[j][i-1]==(1ll<<60)) b[j][i]=1ll<<60;
		else{
			ll S=b[j][i-1]/n,id=b[j][i-1]+1-S*n;
			b[j][i]=min(1ll<<60,b[id][i-1]+S*n);
		}
	ll now=0;
	Rep(i,58,0){
		ll S=now/n,id=now+1-S*n;
		if (b[id][i]+S*n<=k*n)
			now=b[id][i]+S*n;
	}
	For(i,1,N-1) las[i]=0;
	for (now++;now<=k*n;now++){
		ll S=(now-1)/n,id=now-S*n;
		if (!las[a[id]]) q[++*q]=a[id],las[a[id]]=1;
		else{
			for (;q[*q]!=a[id];--*q) las[q[*q]]=0;
			--*q; las[a[id]]=0;
		}
	}
	For(i,1,*q) printf("%d ",q[i]);
	puts("");
}