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;

const int maxn = 5e5 + 10;

int n;
long long k;
int a[maxn], lst[maxn], f[44][maxn];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
    cin >> n >> k;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        a[i + n] = a[i];
    }
    for (int i = 2 * n - 1; i >= n; --i) lst[a[i]] = i;
    for (int i = n - 1; i >= 0; --i) {
        int x = lst[a[i]] + 1;
        if (x >= n) f[0][i] = x - n;
        else f[0][i] = f[0][x];
        lst[a[i]] = i;
    }
    f[0][n] = 0;
    for (int i = 1; i < 44; ++i) for (int j = 0; j <= n; ++j) 
        f[i][j] = f[i - 1][f[i - 1][j]];
    
    int now = 0; k--;
    for (int i = 40; i >= 0; --i) if (k >> i & 1) {
        now = f[i][now];
        k -= 1ll << i;
    }
    memset(lst, 0, sizeof lst);
    vector <int> result;
    for (int i = now; i < n; ++i) {
        if (lst[a[i]] > 0) {
            while (result.size() && lst[a[i]] > 0) {
                lst[result.back()]--;
                result.pop_back();
            }
        } else {
            result.push_back(a[i]);
            lst[a[i]]++;
        }
    }
    for (int it : result) cout << it << " ";
    return 0;
}