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<iomanip>
//#include<cstdio>
#include<vector>
#include<map>
#include<queue>
#include<algorithm>
#include<cmath>
#include<cassert>
using namespace std;
typedef long long ll;
const int Nmax = 2e5;
int N, A[Nmax];
ll K;

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

	vector<int> next(N, -1), pos(Nmax+1, -1);
	for(int i=0; i<N; i++){
		if(pos[A[i]]>=0){
			next[pos[A[i]]] = i;
		}
		pos[A[i]] = i;
	}
	for(int i=0; i<N; i++){
		if(next[pos[A[i]]]<0){
			next[pos[A[i]]] = i;
		}
	}

	int interval = 1;
	int idx = 0;
	while(idx!=N){
		if(next[idx]<=idx) interval++;
		idx = next[idx]+1;
	}

	K %= interval;

	//cout << K << endl;

	idx = 0;
	while(K>1){
		if(next[idx]<=idx) K--;
		idx = next[idx]+1;
		//cout << idx << " " << K << endl;
	}

	vector<int> ans;
	while(idx<N){
		if(next[idx]>idx) idx = next[idx]+1;
		else{
			ans.push_back(A[idx]);
			idx++;
		}
	}

	for(int i=0; i<ans.size(); i++) cout << ans[i] << (i==ans.size()-1?'\n':' ');

	return 0;
}
