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;
const int N = 200005;
int a[N],b[N],h[N],to[N];
int n,w;
ll K;
void solve(int p,int s)
{
	memset(b,0,sizeof(b));
	for(int i = 1;i <= s;i ++)
	{
		if(!b[a[p]])
			h[++ w] = a[p],b[a[p]] = 1;
		else
		{
			while(h[w] != a[p])
				b[h[w]] = 0,w --;
			b[h[w]] = 0;w --;
		}
		p = p%n+1;
	}
	for(int i = 1;i <= w;i ++)
		printf("%d ",h[i]);
}
void solve2(int p,ll s)
{
	while(to[p]-p+1 <= s)
	{
		s -= to[p]-p+1;
		p = to[p]%n+1;
	}
	solve(p,s);
}
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 --)
		b[a[i]] = i+n;
	for(int i = n;i >= 1;i --)
		to[i] = b[a[i]],b[a[i]] = i;
	memset(b,0,sizeof(b));
	int p = 1;ll s = 0,sc = 0;
	while(!b[p])
	{
		b[p] = 1;
		s += to[p]-p+1;
		p = to[p]%n+1;
	}
	int q = p;
	while(true)
	{
		sc += to[q]-q+1;
		q = to[q]%n+1;
		if(q == p)
			break;
	}
	if(n*K <= s)
		solve(1,n*K);
	else
		solve2(p,(n*K-s)%sc);
    return 0;
}