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, A[200000];
    int64_t K;
    cin >> N >> K;
    for(int i=0; i<N; i++) cin >> A[i];

    int nxt[200001], last[200001];
    nxt[N] = N;
    for(int i=0; i<=200000; i++) last[i] = N;
    for(int i=N-1; i>=0; i--){
        nxt[i] = last[A[i]];
        last[A[i]] = i;
    }

    int term[200001];
    term[N] = N;
    for(int i=N-1; i>=0; i--){
        if(nxt[i] == N){
            term[i] = i;
        }else{
            term[i] = term[nxt[i]+1];
        }
    }

    int pos = N, period = 0;
    vector<int> result = {N};
    while(true){
        period++;
        pos = (pos == N ? 0 : last[A[pos]] + 1);
        pos = term[pos];
        if(pos == N) break;
        result.push_back(pos);
    }

    K %= period;
    int start = result[K];
    vector<int> ans;
    set<int> st;
    for(int i=start; i<N; i++){
        if(st.count(A[i])){
            while(st.count(A[i])){
                st.erase(ans.back());
                ans.pop_back();
            }
        }else{
            ans.push_back(A[i]);
            st.insert(A[i]);
        }
    }

    for(int a : ans) cout << a << " ";
    cout << endl;
    return 0;
}
