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
// B.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 2e5 + 200, LOG = 60;

int n, ai[MAX_N], mem[MAX_N], nxt[MAX_N], up[LOG][MAX_N], ansBox[MAX_N];
long long k;

int main()
{
	scanf("%d%lld", &n, &k);
	for (int i = 1; i <= n; i++)
		scanf("%d", &ai[i]);
	for (int i = n; i >= 1; i--)
		mem[ai[i]] = i;
	for (int i = n; i >= 1; i--)
		nxt[i] = mem[ai[i]], mem[ai[i]] = i;
	up[0][0] = 1;
	for (int i = n; i >= 1; i--)
		if (nxt[i] <= i)
			if (nxt[i] == n)
				up[0][i] = 0;
			else
				up[0][i] = nxt[i] + 1;
		else if (nxt[i] == n)
			up[0][i] = 1;
		else
			up[0][i] = up[0][nxt[i] + 1];
	for (int i = 1; i < LOG; i++)
		for (int j = 0; j <= n; j++)
			up[i][j] = up[i - 1][up[i - 1][j]];
	int ans = 0;
	for (int i = LOG - 1; i >= 0; i--)
		if (k & (1LL << i))
			ans = up[i][ans];
	memset(mem, 0, sizeof(mem));
	int tot = 0;
	if (ans == 0)
		return 0;
	for (int i = ans; i <= n; i++)
	{
		if (mem[ai[i]] == 0)
			ansBox[++tot] = ai[i], mem[ai[i]] = tot;
		else
		{
			int pos = mem[ai[i]];
			while (tot >= pos)
				mem[ansBox[tot--]] = 0;
		}
	}
	for (int i = 1; i <= tot; i++)
		printf("%d ", ansBox[i]);
	puts("");
	return 0;
}