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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int nxt[200005][65];
LL sts[200005][65];
LL k;
int n, A[200005];
int l_occ[200005];

int main()
{
	scanf("%d%lld", &n, &k);
	rep(i, n) scanf("%d", &A[i]);
	for(int i = n - 1; i >= 0; i --) l_occ[A[i]] = i;
	for(int i = n - 1; i >= 0; i --) {
		nxt[i][0] = l_occ[A[i]];
		sts[i][0] = l_occ[A[i]] <= i;
		l_occ[A[i]] = i;
	}
	
	rep(i, n) if(nxt[i][0] == n - 1) {
		nxt[i][0] = 0;
		sts[i][0] ++;
	} else nxt[i][0] ++;
	
	rep1(i, 60) rep(j, n) {
		nxt[j][i] = nxt[nxt[j][i - 1]][i - 1];
		sts[j][i] = sts[j][i - 1] + sts[nxt[j][i - 1]][i - 1];
	}
	
	int cur = 0;
	for(int i = 60; i >= 0; i --) if(sts[cur][i] < k) {
		k -= sts[cur][i];
		cur = nxt[cur][i];
	}
	
	for(int i = cur; i < n; i ++)
	if(sts[i][0] == (nxt[i][0] == 0)) {
		if(nxt[i][0] == 0) break;
		i = nxt[i][0] - 1;
	} else printf("%d ", A[i]);
	return 0;
}