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()
{
	cin>>N>>K; 
	for(int i = 0; i < N; i++)
		cin>>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;//第二次出现的位置 
		prv[A[i % N]] = i % N;//第一次出现的位置 
	}
	vector <int> ans;
	memset(vis, -1, sizeof vis);
	int j = 0, k = 0;
	while(k < K)
	{
		//如果就在后面，不需要再来一遍 
		if(nxt[j] > j)
		{
			//直接变过去 
			j = nxt[j] + 1;
			if(j >= N)//超过长度了，次数记得加以，重新循环 
			{
				j -= N;
				k++;
			}
		}
		else//说明到了 下一次循环 
		{
			if(k + 1 < K)//先判断 是否要退出 ，先进行不会退出的 
			{
				j = nxt[j] + 1;//退出当前环 
				k++;//循环次数++ 
				if(j >= N)//这个时候也能要进行下一次循环 
				{
					j -= N;
					k++;
				}
			}
			else//说明要退出 
			{
				ans.push_back(A[j]);
				j = j + 1;//这个时候要一点一点往下推进 
				if(j >= N)
					break;
			}
		}
		//判断这个 位置 是否出现过，作为循环的起点 ，如果出现过的话，就是一次彻底的循环 
		if(vis[j] != -1)
		{
			int d = k - vis[j];//此时的循环 - 出现时的循环次数 
			K = k + (K - k) % d;//还需要的次数 
		}
		vis[j] = k;//记录第一次出现时的循环次数 
	}
	for(auto x : ans)
		cout<<x<<" "; 
	return 0;
}
