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;
#define REP(i,n) for(int i=0; i<(n); ++i)

int N;
long long K;
vector<int> A, B;
vector<int> dp, nxt, same[200200];

void print_result(int piv) {
    if (piv == -1) return;
    while (piv < N) {
        if (nxt[piv] <= N) {
            piv = nxt[piv];
            continue;
        } else {
            cout << A[piv++] << endl;
        }
    }
}

int main() {

    cin >> N >> K;

    A.resize(N);
    REP(i, N) cin >> A[i];

    // B : Aを2回繰り返した配列
    REP(i, 2 * N) B.emplace_back(A[i % N]);

    // same[i] : 数iが格納されているindexのvector
    REP(i, 2 * N) same[B[i]].emplace_back(i);

    // nxt[i] : Bの中で, B[i]と同じ数が次に現れるindexの次のindex
    nxt.resize(N);
    REP(i, N) nxt[i] = *upper_bound(same[B[i]].begin(), same[B[i]].end(), i) + 1;

    int cur = 0;
    while (true) {
        if (nxt[cur] > N) {         // 数列を跨ぐとき
            dp.emplace_back(cur);
        }
        if (nxt[cur] % N == 0) {    // 1巡終了
            dp.emplace_back(-1);    // 数列sが空
            break;
        }
        cur = nxt[cur] % N;
    }

    print_result(dp[(K - 1) % (int)dp.size()]);

}
