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;
#define ll long long

vector<set<int>> Nu(200009);
vector<pair<int,int>> T(200009);
vector<int> ordl(200009);
vector<int> B(200009, -1);
ll tu = 1, N, K, a, ans;

void runner(int x) {
	//xは出発点
	//yは到着点
	int y = T[x].first;
	//cout << x << " " << y << endl;
	if (T[x].second)
		tu++;

	if (tu == K + 1) {
		tu--;
		return;
	}
	if (y == N - 1) {
		K %= tu;
		return;
	}
	B[tu] = y;
	runner(y + 1);
}

int main() {
	cin >> N >> K;
	vector<int> A(N);
	for (int i = 0;i < N;i++) {
		cin >> A[i];
		Nu[A[i]].insert(i);
	}
	for (int i = 0;i < N;i++) {
		auto it = Nu[A[i]].find(i);
		it++;
		if (it == Nu[A[i]].end()) {
			it = Nu[A[i]].begin();
			T[i] = { *it,1 };
		}
		else
			T[i] = { *it,0 };
	}
	B[0] = N;
	runner(0);
	//cout << tu << K << B[K] << endl;
	for (int i = B[K] + 1;i < N;i++) {
		if (T[i].second == 0) {
			i = T[i].first;
		}
		else {
			cout << A[i] << " ";
		}
	}
}