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: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

#define int long long

using ll = long long;
const int N = (int)2e5 + 5, LOG = 63;
int n, a[N], nxt[N][LOG];
ll cost[N][LOG];
bool ex[N];
map<int, int> lst;
vector<int> ans;
ll K;

int32_t main () {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    cin >> n >> K; --K;
    for (int i = 1; i <= n; ++i) cin >> a[i];

    for (int i = n; i > 0; --i) {
        if (lst.count(a[i]) ) nxt[i][0] = lst[ a[i] ];
        else nxt[i][0] = -1;
        lst[ a[i] ] = i;
    }

    for (int i = 1; i <= n; ++i) {
        if (nxt[i][0] == -1) nxt[i][0] = lst[ a[i] ];
        if (nxt[i][0] <= i || nxt[i][0] == n) cost[i][0] = 1;
        nxt[i][0] = nxt[i][0] % n + 1;
    }

    for (int j = 1; j < LOG; ++j) {
        for (int i = 1; i <= n; ++i) {
            nxt[i][j] = nxt[ nxt[i][j - 1] ][j - 1];
            cost[i][j] = cost[i][j - 1] + cost[ nxt[i][j - 1] ][j - 1];
        }
    }

    int cur = 1;
    for (int i = LOG - 1; i >= 0; --i) if (cost[cur][i] <= K) {
        K -= cost[cur][i];
        cur = nxt[cur][i];
    }

    for (int i = cur; i <= n; ++i) {
        if (ex[ a[i] ]) {
            while (ans.back() != a[i]) {
                ex[ans.back()] = 0;
                ans.pop_back();
            }
            ex[ans.back()] = 0;
            ans.pop_back();
        }
        else {
            ans.push_back(a[i]);
            ex[ a[i] ] = 1;
        }
    }

    for (int i = 0; i < (int)ans.size(); ++i) cout << ans[i] << ' '; cout << '\n';

    return 0;
}