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 <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
set<int> s;
int a[200010];
int dp[200010];
vector<int> v[200010];
int main(){
	long long i,n,k;
	cin >> n >> k;
	for(i=0;i<n;i++){
		cin >> a[i];
		v[a[i]].push_back(i);
	}
	dp[n] = 0;
	for(i=n - 1;i>=0;i--){
		int x = a[i];
		if(v[x].back()==i){
			dp[i] = x;
		}else{
			int y = upper_bound(v[x].begin(),v[x].end(),i) - v[x].begin();
			dp[i] = dp[v[x][y] + 1];
		}
	}
	int l = 0,r = n;
	for(i=1;i<=n + 1;i++){
		if(l==0) l = dp[0];
		else l = dp[v[l][0] + 1];
		if(l==0){
			r = i;
			break;
		}
	}
	k = k%r;
	l = 0;
	for(i=1;i<k;i++){
		if(l==0) l = dp[0];
		else l = dp[v[l][0] + 1];
		if(l==0){
			r = i;
			break;
		}
	}
	vector<int> w;
	if(l!=0){
		w.push_back(l);
		s.insert(l);
	}
	for(i=0;i<n;i++){
		if(s.count(a[i])){
			while(1){
				int x = w.back();
				s.erase(x); w.pop_back();
				if(x==a[i]) break;
			}
		}else{
			s.insert(a[i]);
			w.push_back(a[i]);
		}
	}
	for(i=0;i<w.size();i++){
		cout << w[i] << " ";
	}
	cout << endl;
}
	