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<cstdio>
#include<cstring>
using namespace std;
#define N 200020
typedef long long ll;
inline ll read(){
	ll x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N],pre[N],nxt[N],st[N],top,vis[N];
ll k;
int main(){
	n=read(),k=read();
	for(int i=1;i<=n;++i){
		a[i]=read();
	}
	for(int i=1;i<=n;++i){
		if(pre[a[i]])nxt[pre[a[i]]]=i;
		pre[a[i]]=i;
	}
	for(int i=1;i<=n;++i){
		if(!nxt[pre[a[i]]])nxt[pre[a[i]]]=i;
	}
	int t=1;
	for(int u=1;u<=n;u=nxt[u]+1){
		if(nxt[u]<=u)++t;
	}
	k%=t;
	if(!k)return 0;
	for(int u=1,nt=1;u<=n;u=nxt[u]+1){
		if(nxt[u]<=u)++nt;
		if(nt>k){
			for(int i=u;i<=n;++i){
				if(vis[a[i]]){
					while(vis[a[i]])vis[st[top--]]=0;
				}
				else{
					vis[st[++top]=a[i]]=1;
				}
			}
			break;
		}
	}
	for(int i=1;i<=top;++i){
		printf("%d ",st[i]);
	}
	printf("\n");
	return 0;
}


