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"
#include<cassert>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
const long long mod = 1000000007;
const long long inf = 1ll << 61;
typedef pair<int, int> P;
int a[200006];
vector<int>A[200006];
signed main() {
	int n, k; cin >> n >> k;
	rep(i, n)cin >> a[i];
	rep(i, n)A[a[i]].push_back(i);
	int ns = 1;
	int now = a[0];
	int nk = 0;
	while (1) {
		auto x = upper_bound(A[now].begin(), A[now].end(),nk );
		if (x == A[now].end()) {
			ns++;
			nk=A[now][0]+1;
			now = a[nk];
		}
		else {
			nk = *x+1;
			now = a[nk];
		}
		if (nk == n)break;
	}
	k %= ns;
	vector<int>ans;
	ns = 1;
	now = a[0];
	nk = 0;
	while (1) {
		auto x = upper_bound(A[now].begin(), A[now].end(), nk);
		if (x == A[now].end()) {
			if (ns == k)break;
			else ns++;
			nk = A[now][0] + 1;
			now = a[nk];
		}
		else {
			nk = *x + 1;
			now = a[nk];
		}
		if (nk == n)break;
	}
	for (int i = nk; i < n; i++) {
		now = a[i];
		auto x = upper_bound(A[now].begin(), A[now].end(), i);
		if (x == A[now].end()) {
			ans.push_back(now);
		}
		else {
			i = *x;
		}
	}
	rep(i, ans.size()) {
		if (i)cout << ' ';
		cout << ans[i];
	}cout << endl;
}