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 ll long long
#define pii pair<int,int>
using namespace std;
const int mod=1e9+7;
int m,yy[200005],loc[200005],nex[200005],cnt=1,fir[200005],vis[200005],ans[200005]; ll n;

int main()
{
    std::ios::sync_with_stdio(0); std::cin.tie(0);
	cin>>m>>n;
	for(int i=1;i<=m;++i)
	{
		cin>>yy[i];
		if(loc[yy[i]]) nex[loc[yy[i]]]=i;
		else fir[yy[i]]=i;
		loc[yy[i]]=i;
	}
	for(int i=1;i<=m;++i)
	{
		if(vis[i]!=0) continue;
		if(!nex[i]) vis[i]=i;
		else
		{
			int now=i;
			while(nex[now]) loc[++loc[0]]=now,now=nex[now]+1;
			if(now>m) now=-1; 
			for(int j=1;j<=loc[0];++j) vis[loc[j]]=now;
			loc[0]=0;
		}
	}
	int now=1;
	while(1)
	{
		ans[cnt++]=vis[now];
        now=fir[yy[vis[now]]]+1;
        if(vis[now]==-1||now>m) break; 
	}
	now=n%cnt;
	if(now!=0)
	{
		now=ans[now];
		while(now<=m)
		{
			if(!nex[now]) { cout<<yy[now]<<" "; now++;}
			else now=nex[now]+1;
		}
	}
	return 0;
}
