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<vector>
using namespace std;
int N;
long K;
int A[2<<17];
int to[2<<17];
int nxt[2<<17];
long ap[2<<17];
int main()
{
	cin>>N>>K;
	int cnt=0;
	for(int i=0;i<2<<17;i++)nxt[i]=ap[i]=-1;
	for(int i=0;i<N;i++)cin>>A[i];
	for(int i=2*N;i--;)
	{
		to[i%N]=nxt[A[i%N]];
		if(nxt[A[i%N]]==-1)cnt++;
		nxt[A[i%N]]=i;
	}
	int id=0,tern=0;
	ap[id]=K;
	while(K>1)
	{
		int xt=to[id];
		if(K<=(xt+1)/N)
		{
			if((xt+1)/N==2)
			{
				K=0;
			}
			break;
		}
		K-=(xt+1)/N;
		id=(xt+1)%N;
		if(ap[id]>=0)K%=ap[id]-K;
		//if(ap[id]>=0&&K>2)K-=(K-2)/(tern-ap[id])*(tern-ap[id]);
		ap[id]=K;
	}
	if(K==0)
	{
		cout<<endl;
		return 0;
	}
	for(int i=0;i<2<<17;i++)ap[i]=0;
	vector<int>ans;
	for(int i=id;i<N;i++)
	{
		if(ap[A[i]]==0)
		{
			ans.push_back(A[i]);
			ap[A[i]]++;
		}
		else
		{
			while(ap[A[i]]>0)
			{
				ap[ans.back()]--;
				ans.pop_back();
			}
		}
	}
	for(int i=0;i<ans.size();i++)cout<<ans[i]<<" ";
	cout<<endl;
}
