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>
using namespace std;
int P[200005];
int tmp[200005];
int tmp2[200005];
int A[200005];
long long N, K;
int gh[200005];
#define MX 200000

void perm(){
    int* lastpos = tmp;
    int* firstpos = tmp2;
    fill(lastpos, lastpos+200005, -1);
    fill(firstpos, firstpos+200005, -1);

    for (int i=0, k; i<N; i++){
        k = A[i];
        if (lastpos[k] != -1)
            P[lastpos[k]] = (i+1)%N;
        if (firstpos[k] == -1)
            firstpos[k] = i;
        lastpos[k] = i;
        P[i] = firstpos[k] + 1;
    }
}

int calc_cycle(){
    int i=0;
    int cycle=0;
    do {
//        cout << "At " << i << endl;
        if (P[i] <= i + 1)
            cycle++, gh[cycle] = i;
        i = P[i];
    } while (i != 0);
    return cycle;
}

int main(){
    cin >> N >> K;
    for (int i=0;i<N; i++)
        cin >> A[i];
    perm();
    int cycle = calc_cycle();
    K%=cycle;
    int i = gh[K];
    while (i < N){
        if (P[i] == 0)
            break;
        if (P[i] > i+1)
            i = P[i];
        else
            cout << A[i++] << ' ';
    }

}
