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 <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <set>

using ll = long long;

using namespace std;

int main() {
    long n, k;
    cin >> n >> k;
    int a[n], last[200001], next[n];
    fill(last, last+200001, -1);
    fill(next, next+n, 2000000);
    for (auto &ai:a) cin >> ai;
    for (int i=0; i<2*n; i++) {
        int ai = a[i % n];
        if (0 <= last[ai] && last[ai] < n && next[last[ai]] > i) next[last[ai]] = i + 1;
        last[ai] = i;
    }
    int cur = 0;
    long cnt = 0;
    set<long> starts;
    while (true) {
        starts.insert(cnt);
        cnt += next[cur] - cur;
        cur = next[cur] % n;
        if (cur == 0) break;
    }
    auto st = prev(starts.upper_bound((n * k) % cnt));
    for (int i=(*st % n); i<n; ) {
        if (next[i] <= n) {
            i = next[i];
            continue;
        }
        cout << a[i] << ' ';
        i++;
    }
    cout << endl;
    return 0;
}