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
//#pragma GCC optimize("O3")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
//#pragma GCC optimize ("unroll-loops")
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using ld = long double;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n;
    ll k;
    cin >> n >> k;
    vector<int> a(n * 2);
    vector<int> nxt(n * 2, -1);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        a[n + i] = a[i];
    }
    map<int, int> pos;
    for (int i = n * 2 - 1; i >= 0; i--) {
        if (pos.count(a[i])) {
            nxt[i] = pos[a[i]];
        }
        pos[a[i]] = i;
    }
    vector<int> suf(n * 2);
    int per = 1;
    int it = 0;
    do {
        it = nxt[it] + 1;
        if (it > n) {
            per++;
        }
        it %= n;
        suf[per] = it;
    } while(it);
    if (k == per) {
        return 0;
    }
    for (int i = suf[k % per]; i < n; i++) {
        if (nxt[i] >= n) {
            cout << a[i] << ' ';
        } else {
            i = nxt[i];
        }
    }
    return 0;
}