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;
#define ll long long
int n;ll k;
int a[200010],head[200010],nxt[200010],ne[200010],cz[200010],t[200010];
int sta[200010],top;bool b[200010];
int calc(int pl){
	for(int i=pl;i<=n;i++){
		if(!b[a[i]])sta[++top]=a[i],b[a[i]]=1;
		else {
			b[a[i]]=0;
			while(sta[top]!=a[i])b[sta[top]]=0,top--;
			top--;
		}
	}
	for(int i=1;i<=top;i++)cout<<sta[i]<<' ';exit(0);
}
int main(){
	cin>>n>>k;
	if(n==1&((k+1)&1))return 0;
	for(int i=1;i<=n;i++)cin>>a[i];
	if(k==1)calc(1);
	for(int i=n;i>=1;i--){
		nxt[i]=head[a[i]];
		head[a[i]]=i;
	}
	for(int i=1;i<=n;i++)if(!nxt[i])nxt[i]=head[a[i]];
	for(int i=n;i>=1;i--)if(nxt[i]<=i)ne[i]=nxt[i];
	else ne[i]=ne[nxt[i]+1];
	for(int i=1;i<=n;i++)ne[i]=ne[i]+1;ne[n+1]=1;
	cz[1]=1;t[1]=1;
	for(int i=2;i<=k;i++){
		cz[i]=ne[cz[i-1]];
		if(i==k)calc(cz[i]);
		if(t[cz[i]]){
			//xunhuanjie:
			int L=t[cz[i]],R=i-1;
			k=(k-L+1)%(R-L+1);
			if(k==0)k=R-L+1;
			calc(cz[k]);
		}
		t[cz[i]]=i;
	}
}
