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 int long long
using namespace std;

int res,poss;
int n,k,a[200010],dist[200010],nxt[200010];
vector<int> v[200010];
int last[200010];

inline void dfs(int pos,int lvl,bool flag)
{
	if(!flag){
		if(dist[pos]!=-1){
			res=lvl;
			poss=pos;
			return;
		}
		dist[pos]=lvl;
		int u=nxt[pos];
		if(u<=pos){last[lvl]=pos;dfs(u,lvl+1,1);}
		else dfs(u,lvl,1);
	}
	else{
		pos++;
		if(pos<=n){
			dfs(pos,lvl,0);
		}
		else dfs(1,lvl+1,0);
	}
}

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n>>k;
	memset(dist,-1,sizeof(dist));
	memset(last,-1,sizeof(last));
	for(int i=1;i<=n;i++) cin>>a[i];
	for(int i=1;i<=n;i++) v[a[i]].push_back(i);
	for(int i=1;i<=2e5;i++) if(!v[i].empty()){
		for(int j=0;j<v[i].size()-1;j++){
			nxt[v[i][j]]=v[i][j+1];
		}
		nxt[v[i][v[i].size()-1]]=v[i][0];
	}
	dfs(1,1,0);
	int v1=dist[poss]-1,v2=res-dist[poss],st=0;
	if(k<=v1){
		if(last[k]==-1){
			st=n+1;
		}
		else{
			st=last[k];
		}
	}
	else{
		k-=v1;
		k=(k-1)%v2+1;
		k+=v1;
		if(last[k]==-1){
			st=n+1;
		}
		else{
			st=last[k];
		}
	}
	vector<int> ans;
	for(int i=st;i<=n;i++){
		if(nxt[i]<=i){
			ans.push_back(a[i]);
		}
		else{
			i=nxt[i];
		}
	}
	for(int i=0;i<ans.size();i++) cout<<ans[i]<<" ";
	cout<<endl;
	return 0;
}