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;
typedef long long ll;
const int MAXN = 200005;

int A[MAXN];
ll nxt[MAXN][64];
vector<int> p[MAXN];
bitset<MAXN> inv;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	ll N, K;
	cin >> N >> K;
	for (int i = 0; i < N; ++i) cin >> A[i];

	for (int i = 0; i < N; ++i) p[A[i]].push_back(i);
	for (int i = 0; i < MAXN; ++i) {
		if (!p[i].empty()) {
			for (int j = 0; j < p[i].size() - 1; ++j) nxt[p[i][j]][0] = p[i][j + 1] + 1 - p[i][j];
			nxt[p[i].back()][0] = p[i].front() + N + 1 - p[i].back();
		}
	}

	for (int j = 0; j < 63; ++j) {
		for (int i = 0; i < N; ++i) {
			if (nxt[i][j] <= N * K) nxt[i][j + 1] = nxt[i][j] + nxt[(i + nxt[i][j]) % N][j];
			else nxt[i][j + 1] = N * K + 1;
		}
	}

	ll x = 0;
	int cnt = 0;
	while (1) {
		int l = 0, h = 63;
		while (l < h) {
			int m = (l + h) >> 1;
			x + nxt[x % N][m] >= N * K ? h = m : l = m + 1;
		}
		if (h <= 0) break;
		x += nxt[x % N][h - 1];
	}

	vector<int> v;
	for (int i = x % N; i < N; ++i) {
		if (inv[A[i]]) {
			while (v.back() != A[i]) {
				inv[v.back()] = 0;
				v.pop_back();
			}
			inv[A[i]] = 0;
			v.pop_back();
		}
		else {
			inv[A[i]] = 1;
			v.push_back(A[i]);
		}
	}
	for (auto& i : v) cout << i << ' ';
}