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: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef long double ld;

const int inf = 1e9;
const ll inf64 = 1e18;

const int LOG = 60;
const int N = 2e5 + 5;

ll m;
int n;
int a[N], where[N];
vector<int> poses[N];
int link[LOG][N];
ll cost[LOG][N];
int used[N];

int main() {

#ifdef debug
    freopen("input.txt", "r", stdin);
#endif

    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> m;

    for (int i = 0; i < n; i++) {
        cin >> a[i];
        where[i] = (int) poses[a[i]].size();
        poses[a[i]].push_back(i);
    }

    for (int i = 0; i < n; i++) {
        int j = where[i];
        j = (j + 1) % (int) poses[a[i]].size();
        j = poses[a[i]][j];
        j = (j + 1) % n;
        link[0][i] = j;
        if ((int) poses[a[i]].size() > 1) {
            cost[0][i] = i <= j ? j - i : n - i + j;
        } else {
            cost[0][i] = n + 1;
        }
//        cout << i << " -> " << link[0][i] << " cost = " << cost[0][i] << "\n";
    }

    for (int j = 1; j < LOG; j++) {
        for (int i = 0; i < n; i++) {
            link[j][i] = link[j - 1][link[j - 1][i]];
            cost[j][i] = min(inf64, cost[j - 1][i] + cost[j - 1][link[j - 1][i]]);
        }
    }

//    for (int j = 0; j <= 2; j++) {
//        cout << j << " : ";
//        for (int i = 0; i < n; i++) {
//            cout << link[j][i] << " ";
//        }
//        cout << "\n";
//    }

    int v = 0;
    ll dist = m * n;

    for (int j = LOG - 1; j >= 0; j--) {
        if (cost[j][v] > dist) continue;
        dist -= cost[j][v];
        v = link[j][v];
    }

//    cout << "v = " << v << "\n";

    vector<int> res;

    if (dist == 0) {v = n;}
  
    for (int i = v; i < n; i++) {
        int x = a[i];
        if (used[x]) {
            while (1) {
                int y = res.back();
                used[y] = 0;
                res.pop_back();
                if (x == y) break;
            }
        } else {
            used[x] = 1;
            res.push_back(x);
        }
    }

    for (int x : res) {
        cout << x << " ";
    }
    cout << "\n";

    return 0;
}