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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,a[400040],last[200020],len[400040];
int st[200020],tp,bg,l,s;
int vis[200020];
bool bj[200020];
long long k;
int main()
{
	scanf("%d %lld",&n,&k);
	for(int i=1;i<=n;++i)scanf("%d",&a[i]),a[i+n]=a[i];
	for(int i=2*n;i;--i)
	{
		if(last[a[i]])len[i]=len[last[a[i]]+1];
		else len[i]=2*n-i+1;
		last[a[i]]=i;
	}
	st[++tp]=len[n+1];
	vis[len[n+1]]=1;
	int x=len[n+1];
	for(int i=2;;++i)
	{
		x=len[n-x+1];
		if(vis[x])
		{
			bg=vis[x];
			l=i-vis[x];
			break;
		}
		vis[x]=i;
		st[++tp]=x;
	}
//	for(int i=1;i<=tp;++i)printf("%d ",st[i]);printf("\n");
	if(k<bg)s=st[k];
	else s=st[(k-bg)%l+bg];
	tp=0;
//	printf("%d\n",s);
	for(int i=n-s+1;i<=n;++i)
	{
//		printf("i:%d %d\n",i,a[i]);
		if(!bj[a[i]])bj[a[i]]=1,st[++tp]=a[i];
		else while(bj[a[i]])bj[st[tp--]]=0;
	}
	for(int i=1;i<=tp;++i)printf("%d ",st[i]);
	return 0;
}
