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 REP(i, n) for(int (i) = 0; (i) < (n); (i)++)
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

typedef long long ll;

int main(void){
	ll n, k;
	cin >> n >> k;

	vector< vector<int> > v(200001);
	vector<int> a(n);

	REP(i, n){
		cin >> a[i];
		v[a[i]].push_back(i);
	}

	ll pos = 0, cnt = 1;
	vector<int> ans;
	while(1){
		//cout << pos << ' ' << cnt << endl;
		int num = a[pos];
		bool flg = true;
		REP(i, v[num].size()){
			if(v[num][i] > pos){
				pos = v[num][i]+1;
				flg = false;
				break;
			}
		}
		if(flg){
			if(cnt < k){
				pos = v[num][0]+1;
				cnt++;
			}else{
				ans.push_back(a[pos]);
				pos++;
			}
		}

		if(pos == n) cnt *= k/cnt;
		if(pos == n && cnt == k) break;
		if(pos == n){
			pos = 0;
			cnt++;
		}
	}

	REP(i, ans.size()){
		cout << ans[i] << ' ';
	}
	cout << endl;

	return 0;
}
