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 <iostream>
#include <vector>
using namespace std;

int a[200009], mn[200009], r[200009];
bool vs[200009], avs[200009];
vector<int> v, ans;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int n; long long k; cin >> n >> k;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for (int i = n; i >= 1; i--) {
		if (mn[a[i]] == 0) r[i] = a[i];
		else r[i] = r[mn[a[i]] + 1];
		mn[a[i]] = i;
	}
	int p = 1;
	while (true) {
		v.push_back(p);
		if (r[p] == 0) break;
		vs[p] = true;
		p = mn[r[p]] + 1;
		if (vs[p]) break;
	}
	k = (k - 1LL) % (long long)v.size();
	p = v[k];
	for (int i = p; i <= n; i++) {
		if (avs[a[i]]) {
			while (ans.back() != a[i]) {
				avs[ans.back()] = false;
				ans.pop_back();
			}
			avs[a[i]] = false;
			ans.pop_back();
		}
		else {
			ans.push_back(a[i]);
			avs[a[i]] = true;
		}
	}
	for (int i = 0; i < ans.size(); i++)
		cout << ans[i] << " ";
	cout << '\n';
	return 0;
}