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 N 200005
using namespace std;
inline ll read(){
	ll x=0; bool f=1; char c=getchar();
	for(;!isdigit(c); c=getchar()) if(c=='-') f=0;
	for(; isdigit(c); c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
	if(f) return x;
	return 0-x;
}
int n,a[N<<1],nxt[N],to[N];
ll k;
int check(int x){return x=(x-1)%n+1;}
int main(){
	n=read(), k=read(); int nn=n<<1;
	for(int i=1; i<=n; ++i) a[i]=read();
	for(int i=n+1; i<=nn; ++i) a[i]=a[i-n];
	for(int i=nn; i>n; --i) nxt[a[i]]=i;
	for(int i=n; i>=1; --i){
		if(nxt[a[i]]) to[i]=nxt[a[i]]+1-i;
		nxt[a[i]]=i;
	}
	int round = 1+to[1]>n ? 1 : 0, i=check(1+to[1]);
	for(; i!=1; i=check(i+to[i])) if(i+to[i]>n) round+=(i+to[i]-1)/n;
	(--k)%=round, ++k;
	round = 1+to[1]>n ? 1 : 0;
	if(round!=k){
		i=check(1+to[1]);
		for(; ; i=check(i+to[i])){
			if(i+to[i]>n) ++round;
			if(round==k) break;
		}
	}
	int inside=0;
	for(; i<=n; ++i){
		if(i+to[i]<=n+1 && inside<=i) inside=i+to[i];
		if(i>=inside) printf("%d ",a[i]);
	}
	return 0;
}
/*
4 2
1 3 1 2
*/