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 - Do Not Duplicate
#include <bits/stdc++.h>

using namespace std;

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);

	long long n, k;
	cin >> n >> k;
	vector<int> a(n);
	for (int i = 0; i < n; i++) cin >> a[i];

	vector<int> p(200000 + 1, -1);
	vector<long long> next(n);
	for (int i = 0; i < n * 2; i++) {
		if (p[a[i % n]] >= 0) next[p[a[i % n]] % n] = i - p[a[i % n]];
		p[a[i % n]] = i;
	}

	long long x = 0;
	for ( ; ; ) {
		x += next[x % n] + 1;
		if (x % n == 0) break;
	}

	long long m = x / n, r = k % m * n;

	for (x = 0; x + next[x % n] < r; )
		x += next[x % n] + 1;
	vector<int> ans;
	p.assign(p.size(), 0);
	for (long long i = x; i < r; i++) {
		long long j = i % n;
		if (ans.size() > p[a[j]] && ans[p[a[j]]] == a[j])
			ans.resize(p[a[j]]);
		else
			ans.push_back(a[j]), p[a[j]] = ans.size() - 1;
	}

	for (int i = 0; i < ans.size(); i++) cout << (i ? " " : "") << ans[i];
	cout << endl;

	return 0;
}
