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;
#define ll long long
const int N=200050;
const int L=61;
const ll inf=1e18;
const ll lim=2e17;
int go[N][L];
ll sum[N][L];
int a[N],last[N];
bool in[N];
bool was[N];
int main()
{
	//printf("%lld %lld %lld",(ll)1<<L,lim,((ll)1<<L)-lim);
	ll n,k;
	scanf("%lld %lld",&n,&k);
	for(int i=1;i<=n;i++)
	{
		scanf("%i",&a[i]);
		if(!last[a[i]]) last[a[i]]=i;
	}
	auto nxt=[&](int x)
	{
		return x==n?1:x+1;
	};
	for(int i=n;i>=1;i--)
	{
		go[i][0]=nxt(last[a[i]]);
		if(!was[a[i]] || go[i][0]==1) sum[i][0]=1;
		if(!was[a[i]] && go[i][0]==1) sum[i][0]=2;
		last[a[i]]=i;
		was[a[i]]=1;
	}
	for(int j=1;j<L;j++) for(int i=1;i<=n;i++)
	{
		go[i][j]=go[go[i][j-1]][j-1];
		sum[i][j]=sum[i][j-1]+sum[go[i][j-1]][j-1];
		sum[i][j]=min(sum[i][j],inf);
	}
	int u=1;
	for(int j=L-1;~j;j--)
	{
        if(sum[u][j]<k)
		{
			k-=sum[u][j];
			u=go[u][j];
		}
	}
	vector<int> ans;
	for(int i=u;i<=n;i++)
	{
		if(in[a[i]])
		{
			in[a[i]]=0;
			while(ans.back()!=a[i]) in[ans.back()]=0,ans.pop_back();
			ans.pop_back();
		}
		else ans.push_back(a[i]),in[a[i]]=1;
	}
	for(int i:ans) printf("%i ",i);
	return 0;
}