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<algorithm>
#include<vector>
using namespace std;
int a[200050],f[200050],st[200050];
bool b[200050];
vector<int> p[200050];
int main(){
	int n,i,x=0,y,t,cnt=1,sl=-1;
	long long k;
	scanf("%d%lld",&n,&k);
	for(i=0;i<n;++i){
		scanf("%d",a+i);
		p[a[i]].push_back(i);
	}
	do{
		f[x]=cnt;
		t=lower_bound(p[a[x]].begin(),p[a[x]].end(),x)-p[a[x]].begin();
		if((y=p[a[x]][(t+1)%p[a[x]].size()])<=x)++cnt;
		if(++y==n){++cnt;y=0;}
		x=y;
	}while(!f[x]);
	if(k>=f[x])k=(k-f[x])%(cnt-f[x]);
	else x=0;
	while(k>1){
		t=lower_bound(p[a[x]].begin(),p[a[x]].end(),x)-p[a[x]].begin();
		if((y=p[a[x]][(t+1)%p[a[x]].size()])<=x)--k;
		if(++y==n){--k;y=0;}
		x=y;
	}
	for(i=x;i<n;++i)if(!b[a[i]])b[st[++sl]=a[i]]=1;
	else{
		while(sl>=0&&st[sl]!=a[i])b[st[sl--]]=0;
		--sl;
		b[a[i]]=0;
	}
	if(k)for(i=0;i<n;++i)if(!b[a[i]])b[st[++sl]=a[i]]=1;
	else{
		while(sl>=0&&st[sl]!=a[i])b[st[sl--]]=0;
		--sl;
		b[a[i]]=0;
	}
	for(i=0;i<=sl;++i)printf("%d ",st[i]);
	putchar('\n');
	return 0;
}