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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
// const long double PI = acos(-1);
constexpr int inf = 1e9;
constexpr ll INF = 2e18;
// constexpr ll MOD = 1e9+7;
constexpr ll MOD = 998244353;

ll n,k;
int a[200010];

void input() {
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
}

ll table[200010][100];
ll pre[200010];

void make_table() {

    for (int i = n * 2 - 1; i >= 0; i--) {
        table[i%n][0] = pre[a[i%n]] - i + 1;
        pre[a[i%n]] = i;
    }

    for (int i = 1; i < 64; i++) {
        for (int j = 0; j < n; j++) {
            if (table[j][i-1] == -1 || table[(table[j][i-1] + j) % n][i-1] == -1) {
                table[j][i] = -1;
                continue;
            }
            table[j][i] = table[j][i-1] + table[(table[j][i-1] + j) % n][i-1];
            if (table[j][i] > k) table[j][i] = -1;
        }
    }

}

bool inque[200010] = {};
void solve() {
    make_table();
    ll pos = 0;
    ll left = n * k;
    // for (int i = 0; i < 64; i++) {
    //     for (int j = 0; j < n; j++) {
    //         cout << table[j][i] << " ";
    //     }
    //     cout << endl;
    // }
    while (1) {
        // cout << pos << " " << left << endl;
        bool f = 1;
        for (int i = 63; i >= 0; i--) {
            if (table[pos][i] == -1) continue;
            if (left >= table[pos][i]) {
                left -= table[pos][i];
                pos += table[pos][i];
                pos %= n;
                f = 0;
                break;
            }
        }
        if (f) break;
    }

    deque<int> dq;
    while (left--) {
        if (!inque[a[pos]]) {
            dq.push_back(a[pos]);
            inque[a[pos]] = 1;
        } else {
            inque[a[pos]] = 0;
            while (dq.back() != a[pos]) {
                dq.pop_back();
            }
            dq.pop_back();
        }
        pos++;
        pos %= n;
    }

    while (!dq.empty()) {
        cout << dq.front();
        dq.pop_front();
        if (dq.empty()) {
            cout << endl;
        } else {
            cout << " ";
        }
    }
}

int main(int argc, char *argv[]) {
    input();
    solve();
    return 0;
}