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 mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int A[200010], a[200010], b[200010];
ll N, K, d[200010][71];


int main() {
	cin >> N >> K;
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
	}
	for (int i = N - 1; i >= 0; --i) {
		a[i] = b[A[i]];
		b[A[i]] = i;
	}
	for (int i = N - 1; i >= 0; --i) {
		a[i] = b[A[i]];
		b[A[i]] = i;
	}
	for (int i = 0; i < N; ++i) {
		if (a[i] <= i) {
			a[i] += N;
		}
		d[i][0] = a[i] - i + 1;
	}
	for (int i = 0; i < 59; ++i) {
		for (int j = 0; j < N; ++j) {
			d[j][i + 1] = d[j][i] + d[(j + d[j][i]) % N][i];
			d[j][i + 1] = min(d[j][i + 1], N*K + 1);
		}
	}
	ll t = 0;
	vector<int>v;
	while (t < N*K) {
		for (int i = 59; i >= 0; --i) {
			if (t + d[t%N][i] <= N*K) {
				t += d[t%N][i];
			}
		}
		if (t == N*K)break;
		v.push_back(A[t%N]);
		++t;
	}
	for (int i : v) {
		cout << i << sp;
	}
	cout << endl;
}