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.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>

using namespace std;

typedef long long LL;

int main(int argc, char* argv[]) {
	LL n, k;
	cin >> n >> k;
	vector<int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	vector<int> first(200002, -1), next(200002, -1), vis(n, -1);
	for (int i = n - 1; i >= 0; --i) {
		next[i] = first[a[i]];
		first[a[i]] = i;
	}
	bool skipped = false;
	int pos = 0;
	vector<int> ans;
	for (LL i = 0; i < k; ++i) {
		while (pos < n) {
			if (!skipped && vis[pos] >= 0) {
				skipped = true;
				LL t = i - vis[pos];
				i += (max(0LL, (k - i - 3) / t)) * t;
			} else {
				vis[pos] = i;
			}
			int next_pos = next[pos];
			if (next_pos < 0) {
				if (i == k - 1) {
					ans.push_back(a[pos++]);
				} else {
					pos = first[a[pos]] + n + 1;
				}
			} else {
				pos = next_pos + 1;
			}
		}
		pos -= n;
	}
	for (auto x : ans) {
		cout << " " << x;
	}
	cout << endl;
	return 0;
}
