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>
#define RI register int
#define CI const int&
using namespace std;
const int N=400005;
int n,a[N],nxt[N],lst[N],pos,cur,ans[N],cnt; long long k;
int main()
{
	//freopen("B.in","r",stdin); freopen("B.out","w",stdout);
	RI i; for (scanf("%d%lld",&n,&k),i=1;i<=n;++i)
	scanf("%d",&a[i]),a[n+i]=a[i];
	for (i=(n<<1);i>n;--i) lst[a[i]]=i-n;
	for (i=n;i;--i) nxt[i]=lst[a[i]],lst[a[i]]=i;
	//for (i=1;i<=n;++i) printf("%d ",nxt[i]);
	for (pos=cur=1,i=1;i<=n;++i)
	{
		int pxt=nxt[pos]; if (pxt<=pos) ++cur;
		pos=pxt+1; if (pos==1) { --cur; break; }
	}
	for (pos=1,k%=cur,cur=0;;)
	{
		int pxt=nxt[pos]; if (pxt<=pos) ++cur;
		if (cur==k) break; pos=pxt+1;
	}
	for (;;)
	{
		int pxt=nxt[pos]; if (pxt<=pos) ans[++cnt]=a[pos++];
		else pos=pxt+1; if (pos>n) break;
	}
	for (i=1;i<=cnt;++i) printf("%d ",ans[i]);
	return 0;
}