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 fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long LL;

struct D{
	int i;
	LL k;
};

const int maxn=2e5+5;

int n,a[maxn],occur[maxn],d0;
LL k;
D p[maxn],d[maxn];
bool bz[maxn];

int ans0,ans[maxn];
void Last(int x)
{
	memset(bz,0,sizeof(bz));
	fo(i,x,n) if (!bz[a[i]])
	{
		ans[++ans0]=a[i];
		bz[a[i]]=1;
	} else
	{
		while (bz[a[i]]) bz[ans[ans0--]]=0;
	}
	fo(i,1,ans0) printf("%d ",ans[i]);
}

int main()
{
	scanf("%d %lld",&n,&k);
	fo(i,1,n) scanf("%d",&a[i]);
	
	fd(i,n,1)
	{
		if (occur[a[i]])
		{
			if (occur[a[i]]==n) p[i]=(D){1,1ll}; else p[i]=(D){occur[a[i]]+1,0ll};
		}
		occur[a[i]]=i;
	}
	memset(occur,0,sizeof(occur));
	fo(i,1,n)
	{
		if (!occur[a[i]]) occur[a[i]]=i;
		if (!p[i].i)
		{
			if (occur[a[i]]==n) p[i]=(D){1,2ll}; else p[i]=(D){occur[a[i]]+1,1ll};
		}
	}
	
	D i=(D){1,1};
	memset(occur,0,sizeof(occur));
	for(; i.k<k; i.k+=p[i.i].k, i.i=p[i.i].i)
	{
		d[++d0]=i;
		if (occur[i.i]) break; else occur[i.i]=d0;
	}
	
	if (i.k==k) {Last(i.i); return 0;}
	LL cir=d[d0].k-d[occur[d[d0].i]].k;
	i.k=d[occur[d[d0].i]].k+(k-d[occur[d[d0].i]].k)/cir*cir;
	
	for(; i.k<k; i.k+=p[i.i].k, i.i=p[i.i].i);
	Last(i.i);
}