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 Int int64_t

using namespace std;

int main() {
	Int N, K;
	cin >> N >> K;
	vector<int> A(N);
	vector<vector<int>> Ax(200010);
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
		Ax[A[i]].emplace_back(i);
	}

	vector<int> nxt(N);
	for (int i = 0; i < N; ++i) {
		auto V = Ax[A[i]];
		int idx = upper_bound(V.begin(), V.end(), i) - V.begin();
		nxt[i] = (idx == V.size() ? V[0] : V[idx]);
	}

	Int cnt = 1;
	int now = 0;
	while (nxt[now] != N - 1) {
		if (now >= nxt[now]) { ++cnt; }
		now = nxt[now] + 1;
	}
	if (now == N - 1) { ++cnt; }

	K %= cnt;
	now = 0;
	while (K > 1) {
		if (now >= nxt[now]) { --K; }
		now = (nxt[now] + 1) % N;
	}

	vector<int> ans;
	while (now < N) {
		if (now < nxt[now]) {
			now = nxt[now] + 1;
		} else {
			ans.emplace_back(A[now]);
			now += 1;
		}
	}
	for (int i = 0; i < ans.size(); ++i) {
		cout << ans[i] << (i < ans.size() - 1 ? " " : "\n");
	}

	return 0;
}
