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;
using ll = long long;
using P = pair<ll, ll>;
int main()
{
    ll n, k;
    cin >> n >> k;
    vector<int> a(n);
    vector<set<int>> vs(2*1e5+10);
    for(int i = 0; i < n; i++)
    {
        cin >> a[i];
        vs[a[i]].insert(i);
    }
    int pos = 0;
    int cnt = 1;
    vector<int> last(n+1, 0);
    while(true)
    {
        int num = a[pos];
        if(vs[num].size() != 1)
        {
            auto itr = vs[num].lower_bound(pos+1);
            if(itr == vs[num].end())
            {
                pos = *vs[num].begin();
                cnt++;
            }
            else pos = *itr;
        }
        else cnt++;
        pos++;
        last[cnt] = pos;
        if(pos == n) break;
    }
    vector<int> ans;
    pos = last[k%cnt];
    while(pos < n)
    {
        int num = a[pos];
        auto itr = vs[num].lower_bound(pos+1);
        if(itr == vs[num].end()) ans.push_back(num);
        else pos = *itr;
        pos++;
    }
    for(int a : ans) cout << a << " ";
    cout << endl;
}
