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 = 200005;

int N, A[MAXN], nxt[MAXN], prv[MAXN], vis[MAXN];
long long K;

int main(){	
	scanf("%d %lld", &N, &K);
	for(int i = 0; i < N; i++)
		scanf("%d", &A[i]);
	
	memset(prv, -1, sizeof prv);
	for(int i = 0; i < 2 * N; i++){
		if(prv[A[i % N]] != -1){
			nxt[prv[A[i % N]]] = i % N;
		}
		prv[A[i % N]] = i % N;
	}
	/*
	for(int i = 0; i < N; i++)
		cout << nxt[i] << ' ';
	cout << endl;*/
	
	vector <int> ans;
	memset(vis, -1, sizeof vis);
	
	int j = 0, k = 0;
	while(k < K){
		if(nxt[j] > j){
			j = nxt[j] + 1;
			if(j >= N){
				j -= N;
				k++;
			}
		}else{
			if(k + 1 < K){
				j = nxt[j] + 1;
				k++;
				
				if(j >= N){
					j -= N;
					k++;
				}
			}else{
				ans.push_back(A[j]);
				
				j = j + 1;
				if(j >= N)break;
			}
		}
		
		if(vis[j] != -1){
			int d = k - vis[j];
			K = k + (K - k) % d;
		}
		vis[j] = k;
	}
	
	for(int i = 0; i < (int)ans.size(); i++){
		if(i > 0)printf(" ");
		printf("%d", ans[i]);
	}
	printf("\n");
	
	return 0;
}
