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;
typedef bitset<35> sta;
int n,a[200005],Next[200005],t[200005];
ll k,p;
bool v[200005];
int getNext(int now){
	p++;
	if (Next[now]<=now){
		p+=Next[now]+n-now;
	}else p+=Next[now]-now;
	now=Next[now];
	if (now==n) now=1;
	else now=now+1;
	return now;
}
int main(){
	cin>>n>>k;
	for (int i=1;i<=n;++i){
		scanf("%d",&a[i]);
	}
	for (int i=n;i>=1;--i){
		if (t[a[i]]) Next[i]=t[a[i]];
		t[a[i]]=i;
	}
	memset(t,0,sizeof t);
	for (int i=1;i<=n;++i){
		if (t[a[i]]==0) t[a[i]]=i;
		if (Next[i]==0){
			Next[i]=t[a[i]];
		}
	}
	int now=1;
	while (1){
		if (v[now]){
			k=((k)%p)*(n%p)%p;
			if (k==0) return 0;
			k--; p=0;
			while (1){
				int tt=getNext(now);
				if (p>=k){
					break;
				}
				now=tt;
			}
			break;
		}
		v[now]=1;
		int tt=getNext(now);
		now=tt;
	}
	memset(t,0,sizeof t);
	deque <int> q;
	for (int i=now;i<=n;++i){
		if (t[a[i]]!=0){
			int z;
			do{
				z=q.back(); q.pop_back();
				t[z]=0;
			}while (z!=a[i]);
		}else q.push_back(a[i]),t[a[i]]=1;
	}
	while (!q.empty()){
		cout<<q.front()<<" "; q.pop_front();
	}
	return 0;
}