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;

int main(){
    int n;
    int64_t k;
    cin >> n >> k;
    vector<int> A(n);
    for (auto &a: A) cin >> a;

    map<int, vector<int>> M;

    for (int i = 0; i < n; i++){
        M[A[i]].push_back(i);
    }

    int i = 0;
    int c = 0;
    while (true){
        int a = A[i];
        auto p = upper_bound(M[a].begin(), M[a].end(), i);
        if (p == M[a].end()){
            p = M[a].begin();
            c++;
        }

        i = *p;
        i = (i + 1) % n;

        if (i == 0)
            break;
    }

    c++;
    k %= c;
    i = 0;

    while (true){
        int a = A[i];
        auto p = upper_bound(M[a].begin(), M[a].end(), i);
        if (p == M[a].end()){
            p = M[a].begin();
            k--;
        }

        if (k == 0)
            break;

        i = *p;
        i = (i + 1) % n;
    }

    for (; i < n; i++){
        int a = A[i];
        auto p = upper_bound(M[a].begin(), M[a].end(), i);
        if (p == M[a].end()){
            cout << A[i] << " ";
            continue;
        }
        i = *p;
    }
    cout << endl;

}