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 <stdio.h>
#include <algorithm>
#include <vector>
#include <string>
const long long mod = 1000000007;



const int MAX = 200000;
std::vector<int>o[MAX + 1];

int search(int* d, int t, int n) {
	if (d[0] == t)return 0;
	int r = 0;
	for (int i = 20; i >= 0; i--) {
		int c = r | (1 << i);
		if (c >= n)continue;
		if (d[c] == t)return c;
		if (d[c] < t)r = c;
	}
	return -1;
}

int main() {
	int N;
	long long K;
	scanf("%d %lld", &N, &K);
	int *A = new int[N];
	for (int i = 0; i < N; i++) {
		scanf("%d", A + i);
		o[A[i]].push_back(i);
	}
	int s = 0;
	long long cnt = 1;
	while (1) {
		int c = search(o[A[s]].data(), s, o[A[s]].size());
		int sd = (o[A[s]][(c + 1) % o[A[s]].size()] + 1) % N;
		if ((c + 1) % o[A[s]].size() == 0)cnt++;
		s = sd;
		if (s == 0)break;
	}
	long long kd = K % cnt;
	if (kd == 0)return 0;
	s = 0;
	for (int q = 0; q < kd;) {
		int c = search(o[A[s]].data(), s, o[A[s]].size());
		int sd = (o[A[s]][(c + 1) % o[A[s]].size()] + 1) % N;
		if ((c + 1) % o[A[s]].size() == 0)q++;
		if (q < kd)s = sd;
	}
	for (int i = s; i < N; i++) {
		int c = search(o[A[i]].data(), i, o[A[i]].size());
		if (c == o[A[i]].size() - 1) {
			printf("%d ", A[i]);
		}
		else {
			i = o[A[i]][c + 1];
		}
	}
	return 0;
}