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>

#define SZ(x) (((int)x.size()))

typedef long long ll;
typedef long double ld;

using namespace std;

const int maxn = 200*1000+10;
bool mark[maxn];
int n, a[maxn], f[maxn], t, cur;
ll k;
vector<int> loc[maxn], ans;

int main()
{
	cin >> n >> k;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		loc[a[i]].push_back(i);
	}
	for (int i = 0; i < maxn; i++)
	{
		for (int j = 0; j < SZ(loc[i]); j++)
			f[loc[i][j]] = (loc[i][(j + 1) % SZ(loc[i])] + 1) % n;
	}
	do
	{
		if (f[cur] <= cur + 1)
			t++;
		if (f[cur] == 0 && cur == n - 1)
			t++;
		cur = f[cur];
	} while (cur != 0);
	k %= t;
	while (k > 1)
	{
		if (f[cur] <= cur + 1)
			k--;
		cur = f[cur];
	}
	if (k > 0)
	{
		for (int i = cur; i < n; i++)
		{
			if (!mark[a[i]])
			{
				mark[a[i]] = 1;
				ans.push_back(a[i]);
			}
			else
			{
				mark[a[i]] = 0;
				while (ans.back() != a[i])
				{
					mark[ans.back()] = 0;
					ans.pop_back();
				}
				ans.pop_back();
			}
		}
	}
	for (auto w : ans)
		cout << w << " ";
	return 0;
}
