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>
#define llint long long

using namespace std;

llint n, k;
llint a[200005], b[400005];
llint succ[400005], nx[200005];

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> k; k--;
	for(int i = 0; i < n; i++) cin >> a[i];
	for(int i = 0; i < n; i++) b[i] = b[n+i] = a[i];
	
	for(int i = 1; i < 200005; i++) nx[i] = 2*n;
	for(int i = 2*n-1; i >= 0; i--){
		succ[i] = nx[b[i]];
		nx[b[i]] = i;
	}
	for(int i = 0; i < n; i++) succ[i] %= n;
	
	llint cnt = 0, p = 0;
	do{
		if(p >= succ[p]) cnt++;
		p = succ[p];
		if(p == n-1) cnt++;
		p = (p+1)%n;
	}while(p != 0);
	
	p = 0;
	if(k > 0){
		k %= cnt;
		while(k > 0){
			if(p >= succ[p]) k--;
			p = succ[p];
			if(p == n-1) k--;
			p = (p+1)%n;
		}
	}
	if(k < 0) p = n;
	
	vector<llint> ans; set<llint> S;
	for(int i = p; i < n; i++){
		if(S.count(a[i])){
			while(S.count(a[i])){
				S.erase(ans.back());
				ans.pop_back();
			}
		}
		else{
			ans.push_back(a[i]);
			S.insert(a[i]);
		}
	}
	for(int i = 0; i < ans.size(); i++){
		cout << ans[i];
		if(i < (int)ans.size()-1) cout << " ";
	}
	cout << endl;
	
	return 0;
}