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;

typedef long long ll;

const int MAXN = 6e5 + 5;
const int LOG = 60;

ll K;
int N;
int a[MAXN];
map <int, int> lst;
int nxt[MAXN][LOG];
ll weight[MAXN][LOG];
map <int, bool> in;

void load() {
	scanf("%d%lld", &N, &K);
	for (int i = 0; i < N; i++) {
		scanf("%d", a + i);
		a[i + N] = a[i];
	}
}

void solve() {
	K--;
	for (int i = 3 * N - 1; i >= 0; i--) {
		int tmp = lst[a[i]] + 1;
		weight[i][0] = tmp / N - i / N;
		nxt[i][0] = tmp % N;
		lst[a[i]] = i;
	}
	
	for (int j = 1; j < LOG; j++)
		for (int i = 0; i < N; i++) {
			nxt[i][j] = nxt[nxt[i][j - 1]][j - 1];
			weight[i][j] = weight[i][j - 1] + weight[nxt[i][j - 1]][j - 1];
		}
	
	int pos = 0;
	for (int i = LOG - 1; i >= 0; i--)
		if (weight[pos][i] <= K) {
			K -= weight[pos][i];
			pos = nxt[pos][i];
		}
	
	vector <int> ans;
	for (int i = pos + K; i < N; i++)
		if (in[a[i]]) 
			while (1) {
				int tmp = ans.back();
				in[tmp] = false;
				ans.pop_back();
				if (tmp == a[i])
					break;
			}
		else {
			ans.push_back(a[i]);
			in[a[i]] = true;
		}
	
	for (auto it : ans)
		printf("%d ", it);
	puts("");
}

int main() {
	load();
	solve();
	return 0;
}