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>
#include <algorithm>
using namespace std;

long long N, K, A[1 << 19], P[1 << 19], nex[1 << 19];
long long par[1 << 19][66];
bool used[1 << 19];

int main() {
	cin >> N >> K;
	for (int i = 0; i < N; i++) { cin >> A[i]; A[N + i] = A[i]; }
	for (int i = N * 2 - 1; i >= 0; i--) {
		nex[i] = P[A[i]] + 1;
		P[A[i]] = i;
	}
	for (int i = 0; i < N; i++) par[i][0] = (nex[i] - i);
	for (int i = 1; i < 66; i++) {
		for (int j = 0; j < N; j++) {
			long long v1 = par[j][i - 1];
			long long pos = (j + v1) % N;
			long long v2 = par[pos][i - 1];
			long long v3 = v1 + v2;
			v3 = min(v3, (1LL << 61));
			par[j][i] = v3;
		}
	}

	long long cx = 0, px = N * K;
	for (int i = 65; i >= 0; i--) {
		if (px >= par[cx][i]) {
			px -= par[cx][i];
			cx = (cx + par[cx][i]) % N;
		}
	}

	vector<int> vec;
	for (long long i = N * K - px; i < N * K; i++) {
		if (used[A[i % N]] == false) { vec.push_back(A[i % N]); used[A[i % N]] = true; }
		else {
			while (used[A[i % N]] == true) { used[vec[vec.size() - 1]] = false; vec.pop_back(); }
		}
	}

	for (int i = 0; i < vec.size(); i++) { if (i) printf(" "); printf("%d", vec[i]); } printf("\n");
	return 0;
}