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;

ll P[66][202020], L[66][202020];
ll A[404040], C[202020];
vector <ll> S;
ll n, k;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll i, j, l, p;
	
	cin >> n >> k;
	
	for(i=0; i<n; i++){
		cin >> A[i];
		A[n + i] = A[i];
	}
	
	for(i=n+n-1; i>=0; i--){
		if(i < n){
			L[0][i] = C[A[i]] - i + 1;
			P[0][i] = (C[A[i]] + 1) % n;
		}
		C[A[i]] = i;
	}
	
	for(i=1; i<60; i++){
		for(j=0; j<n; j++){
			L[i][j] = min((ll)1e18, L[i - 1][j] + L[i - 1][P[i - 1][j]]);
			P[i][j] = P[i - 1][P[i - 1][j]];
		}
	}
	
	for(l=0, p=0; l<n*k; ){
		for(i=59; i>=0 && l<n*k; i--){
			if(l + L[i][p] <= n * k){
				l += L[i][p];
				p = P[i][p];
			}
		}
		if(l == n * k) break;
		S.push_back(A[p]);
		l ++; p = (p + 1) % n;
	}
	
	for(ll &t: S) cout << t << " ";
	
	cout << "\n";
	
	return 0;
}