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;

const int MX = 200000;

int a[MX], fst[MX + 1], lst[MX + 1], seen[MX + 1];

int main() {
	int n;
	long long k;
	ignore = scanf("%d %lld", &n, &k);
	for (int i = 0; i < n; i++) ignore = scanf("%d", a + i);
	
	fill(lst, lst + MX + 1, -1);
	
	fst[n] = -1;
	for (int i = n - 1; i >= 0; i--) {
		int x = a[i];
		if (lst[x] == -1) fst[i] = x;
		else fst[i] = fst[lst[x] + 1];
		lst[x] = i;
	}
	
	fill(seen, seen + n + 1, -1);
	
	seen[n] = 0;
	seen[0] = 1;
	
	vector<int> f = {n, 0};
	for (int i = 2; i <= k; i++) {
		int p = f.back();
		if (fst[p] == -1) {
			f.push_back(0);
		}
		else {
			f.push_back(lst[fst[p]] + 1);
		}
		
		p = f.back();
		if (seen[p] != -1) {
			int s = seen[p];
			k = (k - s) % (i - s) + s;
		}
		else seen[p] = i;
	}
	
	vector<int> ans;
	set<int> S;
	for (int x = f[k]; x < n; x++) {
		if (S.insert(a[x]).second) {
			ans.push_back(a[x]);
		}
		else {
			while (S.count(a[x]) == 1) {
				S.erase(ans.back());
				ans.pop_back();
			}
		}
	}
	
	for (int x : ans) printf("%d ", x);
	printf("\n");
	
	return 0;
}
