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;

int64_t dp[60][2000010];
int main() {
	const int64_t INF = 1e18 / 3;
	int64_t n, k; cin >> n >> k;
	int a[n]; for(int i = 0; i < n; ++i) cin >> a[i];
	int pre[200010]; for(int i = 0; i < 200010; ++i) pre[i] = -1;
	// dp[j][i] := iから2^j回遷移したときの消費回数
	for(int i = 0; i < 2 * n; ++i) {
		if(pre[a[i % n]] != -1) {
			dp[0][pre[a[i % n]]] = (i + 1) - pre[a[i % n]];
		}
		pre[a[i % n]] = i;
	}
	for(int j = 1; j < 60; ++j) {
		for(int i = 0; i < n; ++i) {
			dp[j][i] = min(dp[j - 1][i] + dp[j - 1][(i + dp[j - 1][i]) % n], INF);
		}
	}
	int64_t goal = n * k;
	int64_t now = 0;
	while(true) {
		bool flg = true;
		for(int j = 59; j >= 0; --j) {
			if(dp[j][now % n] == INF) continue;
			if(now + dp[j][now % n] < goal) {
				now += dp[j][now % n];
				flg = false;
				break;
			}
		}
		if(flg) break;
	}
	deque<int> ans;
	bool used[200010] = {};
	cerr << now % n << endl;
	for(int i = now % n; i < n; ++i) {
		if(used[a[i]]) {
			while(ans.size()) {
				int t = ans.back();
				ans.pop_back();
				used[t] = false;
				if(t == a[i]) break;
			}
		} else {
			ans.push_back(a[i]);
			used[a[i]] = true;
		}
	}
	for(int i : ans) {
		cout << i << " ";
	}
	cout << "\n";
	return 0;
}