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>
using namespace std;

inline int read()
{
	int ret=0; char c=getchar();
	while(c<48||c>57)c=getchar();
	while(c>=48 && c<=57)ret=ret*10+c-48,c=getchar();
	return ret;
}
#define N 200093
long long k;
int T,f[N],g[N],u,v,st[N],t,n,a[N],h[N],tt,hh[N];
inline void pr(int u)
{
	while(u<=n){
		if(f[u]>u)u=f[u]; else printf("%d ",a[u]);
		++u;
	}
	exit(0);
} 

int main()
{
	n=read(); cin >> k;
	for(int i=1;i<=n;++i) a[i]=read();
	for(int i=1;i<=n;++i) {
		if(!h[a[i]])h[a[i]]=i;
	}
	for(int i=n;i;--i){
		if(!g[a[i]])f[i]=h[a[i]]; else f[i]=g[a[i]];
		g[a[i]]=i;
	}
	
	/*
	u=1;
	while(1){
		v=f[u]+1;if(v>n)v=1;
		if(v<=u+1)++t;
		if(v==1){
			T=t;
			break;
		}
		u=v;
	}
	*/ 
	
	T=1,u=1;
	while(1){
		if(f[u]<=u)++T;
		u=f[u]+1;
		if(u==n+1)break;
	}
	
	u=1; --k; k%=T; if(!k)pr(u);
	while(1){
		v=f[u]+1;if(v>n)v=1;
		if(v<=u+1)--k;
		if(!k)pr(v);
		u=v;
	}
}