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<cmath>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
const int N=400005;
int n;
long long k;
int a[N];
int pos[N],nxt[N];
int f[N][60];
int vis[N];
int main()
{
	scanf("%d%lld",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]),a[i+n]=a[i];
	for(int i=n*2;i>=1;i--)
		nxt[i]=pos[a[i]],pos[a[i]]=i;
	f[0][0]=1;
	for(int i=n;i>=1;i--)
		if(nxt[i]>n)
		{
			if(nxt[i]==n+n) f[i][0]=0;
			else f[i][0]=nxt[i]-n+1;
		}
		else
		{
			if(nxt[i]==n) f[i][0]=1;
			else f[i][0]=f[nxt[i]+1][0];
		}
	for(int i=1;i<=log2(k);i++)
		for(int u=0;u<=n;u++)
			f[u][i]=f[f[u][i-1]][i-1];
	int p=0;
	for(int i=0;i<=log2(k);i++)
		if(k&(1LL<<i)) p=f[p][i];
	stack<int>s;
	for(int i=p;i<=n;i++)
	{
		if(vis[a[i]])
		{
			while(!s.empty()&&s.top()!=a[i]) vis[s.top()]--,s.pop();
			if(!s.empty()) vis[s.top()]--,s.pop();
		}
		else s.push(a[i]),vis[a[i]]++;
	}
	vector<int>res;
	while(!s.empty()) res.push_back(s.top()),s.pop();
	reverse(res.begin(),res.end());
	for(int u:res)
		printf("%d ",u); 
	return 0;
}