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<cstdio>
typedef long long ll;
const int MAXN = 200000;
int A[MAXN + 5], n; ll k;
int vis[MAXN + 5], stk[MAXN + 5], tp;
void get_ans(int x) {
	for(int i=x;i<=n;i++) {
		if( vis[A[i]] ) {
			while( tp && stk[tp] != A[i] )
				vis[stk[tp--]] = false;
			tp--;
		}
		else stk[++tp] = A[i];
		vis[A[i]] ^= 1;
	}
	for(int i=1;i<=tp;i++)
		printf("%d ", stk[i]);
}
int nxt[60][MAXN + 5];
ll dis[60][MAXN + 5];
int adj[MAXN + 5];
int nt(int x) {return x == n ? 1 : x + 1;}
int main() {
	scanf("%d%lld", &n, &k);
	for(int i=1;i<=n;i++) scanf("%d", &A[i]);
	for(int i=n;i>=1;i--) adj[A[i]] = n + i;
	for(int i=n;i>=1;i--) {
		dis[0][i] = adj[A[i]] / n;
		nxt[0][i] = adj[A[i]] % n + 1;
		adj[A[i]] = i;
	}
	for(int i=1;i<60;i++)
		for(int j=1;j<=n;j++) {
			nxt[i][j] = nxt[i-1][nxt[i-1][j]];
			dis[i][j] = dis[i-1][j] + dis[i-1][nxt[i-1][j]];
		}
	int s = 1;
	for(int i=59;i>=0;i--)
		if( dis[i][s] < k )
			k -= dis[i][s], s = nxt[i][s];
	get_ans(s);
}