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;
const int maxn = 200010;
int n, a[maxn], lst[maxn], nxt[maxn], fir[maxn], f[maxn];
ll k; bool used[maxn];
vector<int> b, V;

int main() {
    scanf("%d %lld", &n, &k);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (!fir[a[i]]) fir[a[i]] = i;
    }
    for (int i = n; i; i--) {
        nxt[i] = lst[a[i]], lst[a[i]] = i;
    }
    for (int i = n; i; i--) {
        if (!nxt[i]) f[i] = fir[a[i]] + 1;
        else f[i] = f[nxt[i] + 1];
    }
    b.push_back(1);
    while (b.back()) {
        b.push_back(f[b.back()]);
    }
    b.pop_back();
    int cur = b[(k - 1) % b.size()];
    for (int i = max(1, cur); i <= n; i++) {
        if (!used[a[i]]) {
            V.push_back(a[i]), used[a[i]] = 1;
        } else {
            while (used[a[i]]) used[V.back()] = 0, V.pop_back();
        }
    }
    for (int x : V) printf("%d ", x);
    return 0;
}