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;
int f[200010][41],a[200010],nxt[200010],pos[200010];
int ans[200010],mark[200010];
int i,j,k,n,m,x,y,t;
ll T;
int main(){
	scanf("%d%lld",&n,&T);
	for (i=1;i<=n;i++)scanf("%d",&a[i]);
	for (i=1;i<=n;i++)if (!pos[a[i]])pos[a[i]]=i;
	for (i=n;i>=1;i--)nxt[i]=pos[a[i]]%n,pos[a[i]]=i;f[n+1][0]=1;
	for (i=n;i>=1;i--){
		if (i==n&&nxt[i]==0){f[i][0]=n+1;continue;}
		if (nxt[i]<=i)f[i][0]=nxt[i]+1;else f[i][0]=f[nxt[i]+1][0];
	}
	for (i=1;i<=40;i++)for (j=1;j<=n+1;j++)f[j][i]=f[f[j][i-1]][i-1];
	x=1;T--;
	for (i=40;i>=0;i--)if ((1ll<<i)<=T)
		x=f[x][i],T-=(1ll<<i);
	for (i=x;i<=n;i++){
		if (mark[a[i]]){
			while (ans[ans[0]]!=a[i]){
				mark[ans[ans[0]]]=0;
				ans[0]--;
			}
			mark[a[i]]=0;ans[0]--;
		}
		else ans[++ans[0]]=a[i],mark[a[i]]=1;
	}
	for (i=1;i<=ans[0];i++)printf("%d ",ans[i]);
	return 0;
}
