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;
const int maxn=2001000;
int n,a[maxn],nxt[maxn],flag[maxn];
long long k,f[maxn][100];
vector<int> pos[maxn];
vector<int> ans;
int main()
{
	scanf("%d%lld",&n,&k);
	for(int i=1;i<=n;i++) scanf("%d",a+i),pos[a[i]].push_back(i);
	for(int i=1;i<=200000;i++) for(int j=0;j<(int)pos[i].size();j++)
	{
		if(j+1==(int)pos[i].size()) nxt[pos[i][j]]=pos[i][0]+n-pos[i][j]+1;
		else nxt[pos[i][j]]=pos[i][j+1]-pos[i][j]+1;
	}
	for(int i=1;i<=n;i++) f[i][0]=nxt[i];
	for(int j=1;j<70;j++) for(int i=1;i<=n;i++)
		f[i][j]=min(n*k+1,f[i][j-1]+f[(i+f[i][j-1]-1)%n+1][j-1]);
	long long now=1;
	for(int i=69;i>=0;i--) if(f[(now-1)%n+1][i]+now<=n*k+1) now+=f[(now-1)%n+1][i];
	for(long long i=now;i<=n*k;i++)
	{
		int val=a[(i-1)%n+1];
		if(flag[val])
		{
			while(ans[ans.size()-1]!=val) flag[ans[ans.size()-1]]=0,ans.pop_back();
			ans.pop_back();flag[val]=0;
		}
		else{ans.push_back(val);flag[val]=1;}
	}
	for(int i=0;i<(int)ans.size();i++) cout<<ans[i]<<' ';
}