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;

typedef long long ll;
const int maxn = 2e5;
int n, a[maxn + 3], b[maxn + 3], nxt[maxn + 3];
int k, p, c[maxn + 3], vis[maxn + 3], sz, ans[maxn + 3];
ll m;

void solve(int x) {
	for (int i = 1; i <= maxn; i++) {
		b[i] = 0;
	}
	for (int i = x; i <= n; i++) {
		if (!b[a[i]]) {
			ans[++sz] = a[i];
			b[a[i]] = sz;
		} else {
			int t = sz;
			sz = b[a[i]] - 1;
			for (int j = b[a[i]]; j <= t; j++) {
				b[ans[j]] = 0, ans[j] = 0;
			}
		}
	}
	for (int i = 1; i <= sz; i++) {
		printf("%d%c", ans[i], " \n"[i == sz]);
	}
}

int main() {
	scanf("%d %lld", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	if (m == 1) {
		solve(1);
		return 0;
	}
	for (int i = n; i; i--) {
		b[a[i]] = i;
	}
	for (int i = n; i; i--) {
		nxt[i] = b[a[i]];
		b[a[i]] = i;
		if (nxt[i] > i) {
			nxt[i] = nxt[nxt[i] + 1];
		}
	}
	c[1] = 1;
	vis[1] = 1;
	for (int i = 2; i <= n + 2; i++) {
		if (c[i - 1] == n + 1) {
			c[i] = 1;
		} else {
			c[i] = nxt[c[i - 1]] + 1;
		}
		if (m == i) {
			solve(i);
			return 0;
		}
		if (vis[c[i]]) {
			k = i - vis[c[i]];
			p = vis[c[i]];
			solve(c[(m - p) % k + p]);
			return 0;
		}
		vis[c[i]] = i;
	}
	return 0;
}