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>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,k,sum,v,nxt[200005],i,dp[60][200005],j,a[200005],pq,pos;
const ll inf=1e18;
int main()
{
	scanf("%lld%lld",&n,&k);
	for(i=0;i<n;i++)scanf("%lld",&a[i]);
	for(i=0;i<n;i++)if(nxt[a[i]]==0)nxt[a[i]]=n+i;
	for(j=n-1;j>=0;j--)
	{
		dp[0][j]=nxt[a[j]]-j;
		nxt[a[j]]=j;	
	}	
	for(i=1;i<60;i++)
	{
		for(j=0;j<n;j++)
		{
			pq=dp[i-1][j]+1LL;
			pos=(j+pq)%n;
			pq+=dp[i-1][pos];
			if(pq>inf)dp[i][j]=inf;
			else dp[i][j]=pq;
		}
	}
	//printf("%lld\n",dp[1][0]);
	v=0;sum=n*k-1LL;
	if(sum==0)
	{
		printf("%lld\n",a[v]);
		return 0;
	}
	while(sum)
	{
		for(i=59;i>=0;i--)
		{
			if(dp[i][v]>sum)continue;
			sum-=dp[i][v];
			v=(v+dp[i][v])%n;
			if(sum==0)return 0;
			sum--;
			//printf("%lld %lld %lld\n",sum,v,dp[i][v]);
			v=(v+1LL)%n;
			if(sum==0)
			{
				printf("%lld ",a[v]);
				return 0;
			}
		}
		printf("%lld ",a[v]);
		sum--;
		v=(v+1)%n;
		if(sum==0)
		{
			printf("%lld ",a[v]);
			return 0;
		}
	}
	//printf("%lld ",a[v]);
	return 0;
} 