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 <algorithm>
#include <vector>
#include <string>
#include <cmath>
using namespace std;

int dir[200001];

int main()
{
	long long N,K;cin>>N>>K;
	vector<int>A(N);
	for(int i=0;i<N;i++)cin>>A[i];

	for (int i = 0; i < 200001; i++)dir[i] = -1;
	vector<int>D(N);
	for (int i = 0; i <2*N; i++)
	{
		D[i%N] = dir[A[i%N]];
		dir[A[i % N]] = i%N;
	}
	for (int i = 0; i < 200001; i++)dir[i] = -1;
	vector<int>revD(N);
	for (int i = 2*N-1; i >= 0; i--)
	{
		revD[i % N] = dir[A[i % N]];
		dir[A[i % N]] = i % N;
	}


	int T=0;

	vector<int>B(N);
	int pos = N-1;
	int t=0;
	while (1)
	{
		B[pos] = t;

		if(D[pos] >= pos)t++;

		pos = D[pos];

		if (pos == 0)break;

		pos = pos-1;
		if (pos < 0)
		{
			pos+=N;
			t++;
		}
	}

	T = t+1;
	for(int i=0;i<N;i++)B[i] = (T-B[i])%T;

	//cerr<<T<<endl;

	K%=T;

	//cerr<<K<<endl;

	/*for (int i = 0; i < N; i++)
	{
		cerr<<B[i]<<" ";
	}
	cerr<<endl;*/

	
	int s = 0;
	for(int i=N-1;i>=0;i--)
	{
		if (B[i] == K)
		{
			s = i+1;
			break;
		}
	}
	
	vector<int>ans;
	for (int i = s; i < N; i++)
	{
		if (revD[i] > i)
		{
			i = revD[i];
			continue;
		}

		ans.push_back(A[i]);
	}

	for (int i = 0; i < ans.size(); i++)
	{
		if(i)cout<<" ";
		cout<<ans[i];
	}
	cout<<endl;

	return 0;
}