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 <iostream>
#include <vector>
using namespace std;
typedef long long ll;

int a[200005];
vector<int> v[200005];
int l[200005];
int re[200005];
int cnt[200005];
int c[200005];

int main()
{
    ll n, k;
    cin >> n >> k;
    for(int i = 0; i < n; i++){
        cin >> a[i];
        cnt[a[i]]++;
        re[i] = cnt[a[i]];
        v[a[i]].push_back(i);
    }
    for(int i = 0; i < n; i++){
        l[a[i]] = v[a[i]].size();
    }
    int now = 0;
    int i;
    for(i = 0; ; i++){
        while(l[a[now]] != re[now]){
            now = v[a[now]][re[now]] + 1;
            if(now == n){
                break;
            }
        }
        c[i] = now;
//        cout << i << " " << c[i] << endl;
        if(l[a[now]] == 1 && now == n - 1){
            i++;
            c[i] = n;
            break;
        }
        if(now == n) break;
        now = (v[a[now]][0] + 1) % n;
    }
    int cir = i + 1;
//    cout << cir << endl;
    int p = (k + cir - 1) % cir;
    for(int i = c[p]; i < n; i++){
        while(l[a[i]] != re[i]){
            i = v[a[i]][re[i]] + 1;
            if(i == n){
                break;
            }
        }
        if(i >= n) break;
        cout << a[i] << " ";
    }
    cout << endl;
}