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 mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=2e5+10;
int n;
int A[maxn];
vector<int> pos[maxn];
ll k;
int has[maxn];
vector<int> tmp;
int main()
{
	scanf("%d%lld",&n,&k);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&A[i]);
		pos[A[i]].emplace_back(i);
	}
	int cur=1,p;
	while(true)
	{
		p=lower_bound(pos[A[cur]].begin(),pos[A[cur]].end(),cur)-pos[A[cur]].begin();
		p++;
		if(p>=pos[A[cur]].size())
		{
			tmp.emplace_back(cur);
	//		cout<<cur<<endl;
			p-=pos[A[cur]].size();
			cur=pos[A[cur]][p]+1;
		}
		else
		{
			cur=pos[A[cur]][p]+1;
		}
		if(cur==n+1)break;
	}
	int m=tmp.size();
	k=k%(1ll*(m+1))-1;
	if(k<0)return 0;
	p=tmp[k];
	tmp.clear();
//	cout<<k<<" "<<p<<endl;
	for(int i=p;i<=n;i++)
	{
		if(!has[A[i]])tmp.emplace_back(A[i]),has[A[i]]++;
		else
		{
			while(has[A[i]])
			{
				has[tmp.back()]--;
				tmp.pop_back();
			}
		}
	}
	for(int i=0;i<tmp.size();i++)
		printf("%d ",tmp[i]);
}