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 pb push_back
using namespace std;
long long n , k;
int a[200005];
int p[200005];
int t[200005];
int c[200005];
vector <long long> periods;
int nxt( int j){
return j%n + 1;
}
int main()
{
    cin >> n >> k;
    for(int i = 1 ; i <= n ; i++){
        cin >> a[i];
    }
    for(int i = 1 ; i <= 2*n ; i++){
        int j = (i-1)%n + 1;
        if(c[a[j]] == 0){
            c[a[j]] = j;
        }else{
            p[c[a[j]]] = j;
            t[c[a[j]]] = ( j - c[a[j]] + n - 1)%n + 1;
            c[a[j]] = j;
        }
    }
    for(int i = 1 ;i <= n; i++){
        //cout << p[i] << " " << t[i] << '\n';
    }
    long long period = 0;
    int pos = 1;
    period += t[pos] + 1;
    pos = nxt(p[pos]);
    while(pos!=1){
        period += t[pos] + 1;
        pos = nxt(p[pos]);
    }
    long long rest = (n*k)%period;
    pos = 1;
    while( rest > 0){
        if(rest > t[pos]){
            rest -= t[pos]+1;
            pos = nxt(p[pos]);
        }else{
            cout << a[pos] << " ";
            pos = nxt(pos);
            rest--;
        }
    }
    return 0;
}
