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;
typedef unsigned long long hxtype;
template<class IT>inline void cmax(IT &a,IT b){if(a<b)a=b;}
template<class IT>inline void cmin(IT &a,IT b){if(b<a)a=b;}
const int N=200005;
int q[N],p[N],a[N],f[N],e[N];
int main(){
	int n,i,s,x=1,m;ll k;
	scanf("%d%lld",&n,&k);p[m=n+1]=1;
	for(i=1;i<=n;++i)scanf("%d",&a[i]);
	for(i=n;i   ;--i)f[a[i]]=i;
	for(i=N;i   ;--i)e[i]=m;
	for(i=n;i   ;--i){
		if(e[a[i]]==m)p[i]=f[a[i]]+1;
		else p[i]=p[e[a[i]]+1];
		e[a[i]]=i;
	}
//	for(i=1;i<=m;++i)printf("%d ",p[i]);printf("\n");
	--k;
	while(k){
		if(k&1ll)x=p[x];
		k>>=1;
		for(i=1;i<=m;++i)q[i]=p[p[i]];
		for(i=1;i<=m;++i)p[i]=  q[i] ;
	}
//	printf("x=%d\n",x);
	s=0;
	for(i=N;i   ;--i)e[i]=0;
	for(i=x;i<=n;++i){
		if(e[a[i]]){
			do e[x=f[s--]]=0;while(x!=a[i]);
		}else{
			e[f[++s]=a[i]]=1;
		}
	}
	for(i=1;i<=s;++i)printf("%d ",f[i]);
	return 0;
}