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;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)

const int size = 2e5 + 1;

int main() {
    ll N, K;
    cin >> N >> K;
    vector<int> A(N);
    rep(i, N) cin >> A[i];

    pair<int, int> g[N];
    vector<int> id(size, -1);
    rep(i, 2 * N) {
        int a = A[i % N];
        if (id[a] == -1)
            id[a] = i;
        else {
            g[id[a] % N] = {i % N, (id[a] < N && N <= i)};
            id[a] = i;
        }
    }
    int l = 0, v = 0;
    while (true) {
        l += g[v].second;
        v = g[v].first + 1;
        if (v == N) break;
    }
    
    K %= l + 1;
    l = v = 0;
    if (K == 0) return 0;
    while (true) {
        if (l == K - 1) break;
        l += g[v].second;
        v = g[v].first + 1;
    }

    deque<int> dq;
    while (v < N) {
        if (v < g[v].first)
            v = g[v].first + 1;
        else {
            dq.push_back(A[v]);
            v++;
        }
    }
    while (!dq.empty()) {
        cout << dq.front() << " ";
        dq.pop_front();
    }
    cout << endl;
}
