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 <vector>
using namespace std;
bool ex[200005];
int arr[200005],f[200005];
pair<long long,int> dp[60][200005];
pair<long long,int> merge(pair<long long,int> a,pair<long long,int> b)
{
	return {a.first+b.first,b.second};
}
int main()
{
	int n;
	long long k;
	scanf("%d%lld",&n,&k);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&arr[i]);
		if (!f[arr[i]])
		f[arr[i]]=i;
	}
	for (int i=n;i>0;i--)
	{
		dp[0][i]={0,f[arr[i]]%n+1};
		if (f[arr[i]]<=i)
		dp[0][i].first++;
		if (f[arr[i]]==n)
		dp[0][i].first++;
		f[arr[i]]=i;
	}
	for (int i=1;i<60;i++)
	{
		for (int j=1;j<=n;j++)
		dp[i][j]=merge(dp[i-1][j],dp[i-1][dp[i-1][j].second]);
	}
	pair<long long,int> cur(1,1);
	for (int i=59;i>=0;i--)
	{
		auto tmp=merge(cur,dp[i][cur.second]);
		if (tmp.first<=k)
		cur=tmp;
	}
	vector<int> ans;
	while (cur.first<=k)
	{
		if (ex[arr[cur.second]])
		{
			while (ex[arr[cur.second]])
			{
				ex[ans.back()]=0;
				ans.pop_back();
			}
		}
		else
		{
			ans.push_back(arr[cur.second]);
			ex[arr[cur.second]]=1;
		}
		if (cur.second==n)
		cur.first++;
		cur.second=cur.second%n+1;
	}
	for (int i:ans)
	printf("%d ",i);
}