Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3 2
1 2 3

Sample Output 1

2 3

In this case, X=(1,2,3,1,2,3).
We will perform the operations as follows:

i=0: s does not contain 1, so we add 1 to the end of s, resulting in s=(1).

i=1: s does not contain 2, so we add 2 to the end of s, resulting in s=(1,2).

i=2: s does not contain 3, so we add 3 to the end of s, resulting in s=(1,2,3).

i=3: s does contain 1, so we repeatedly delete the element at the end of s as long as s contains 1, which causes the following changes to s: (1,2,3)→(1,2)→(1)→().

i=4: s does not contain 2, so we add 2 to the end of s, resulting in s=(2).

i=5: s does not contain 3, so we add 3 to the end of s, resulting in s=(2,3).

Sample Input 2

5 10
1 2 3 2 3

Sample Output 2

3

Sample Input 3

6 1000000000000
1 1 2 2 3 3

Sample Output 3

s may be empty in the end.

Sample Input 4

11 97
3 1 4 1 5 9 2 6 5 3 5

Sample Output 4

9 2 6

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;

const int MAX_N = 2 * 1e5 + 2;

bool is_used[MAX_N];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N, K;
    cin >> N >> K;
    vector<int> A(N);
    for (int i = 0; i < N; i++) cin >> A[i];
    vector<ll> last_pos(MAX_N, -1);
    vector<ll> next_pos(N, -1);
    for (ll i = 2 * N - 1; i >= 0; i--) {
        if (last_pos[A[i % N]] >= 0) {
            next_pos[i % N] = last_pos[A[i % N]];
        }
        last_pos[A[i % N]] = i % N;
    }
    ll now_idx = 0, next_idx = next_pos[now_idx];
    ll period = 0;
    if (now_idx < next_idx) period += next_idx - now_idx + 1;
    else if (now_idx > next_idx) period += N - now_idx + next_idx + 1;
    else period += N + 1;
    while (next_idx != N - 1) {
        now_idx = (next_idx + 1) % N;
        next_idx = next_pos[now_idx];
        if (now_idx < next_idx) period += next_idx - now_idx + 1;
        else if (now_idx > next_idx) period += N - now_idx + next_idx + 1;
        else period += N + 1;
    }
    ll rest = (N * K) % period;
    vector<int> s;
    for (ll i = 0; i < rest; i++) {
        if (is_used[A[i % N]]) {
            while (is_used[A[i % N]]) {
                is_used[s[s.size() - 1]] = false;
                s.pop_back();
            }
        }
        else {
            s.push_back(A[i % N]);
            is_used[A[i % N]] = true;
        }
    }
    for (int i = 0; i < s.size(); i++) cout << s[i] << ' ';
    cout << endl;
    return 0;
}