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>
#define ll long long
#define xx first
#define yy second
using namespace std;
const int N = 200000 + 5;
int h[N], suf[N], info[N], n;
int stk[N], top;
ll k;
bool vis[N];

int main () {
	#ifdef wll
	freopen ("testdata.in", "r", stdin);
	#endif
	scanf ("%d%lld", &n, &k);
	for (int i = 1; i <= n; ++i) {scanf ("%d", &info[i]);}
	for (int i = 1; i <= n; ++i) {if (h[info[i]]) {suf[h[info[i]]] = i;}h[info[i]] = i;}
	for (int i = n; i >= 1; --i) {h[info[i]] = i;}
	for (int i = 1; i <= n; ++i) {if (!suf[i]) {suf[i] = h[info[i]];}}
	int cir = 1, now = 1;
	while (1) {
		bool flag = false;
		if (suf[now] <= now) {cir++;flag = true;}
		now = suf[now] % n + 1;
		if (vis[now]) {if (flag)--cir;break;}
		vis[now] = 1;
	}
	k %= cir;
	now = 1;
	int cnt = 1;
	while (1) {
//		printf ("%d %d\n", now, cnt);
		if (now > n) {now -= n;cnt++;}
		if (now > n || cnt > k) {break;}
		int nxt = suf[now];
		if (nxt <= now) {
			if (1LL * cnt >= k) {stk[++top] = info[now++];continue;}
			else {cnt++; now = suf[now] + 1;}
		}
		else {
			now = suf[now] + 1;
		}
	}
	for (int i = 1; i <= top; ++i) {
		printf ("%d ", stk[i]);
	}
	return 0;
}









