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;

using pii = pair<int,int>;
using ll = long long;
using ull = unsigned long long;
#define rep(i, j) for(int i=0; i < (int)(j); i++)
#define repeat(i, j, k) for(int i = (j); i < (int)(k); i++)
#define all(v) v.begin(),v.end()
#define debug(x) cerr << #x << " : " << x << endl

template<class T> bool set_min(T &a, const T &b) { return a > b  ? a = b, true : false; }
template<class T> bool set_max(T &a, const T &b) { return a < b  ? a = b, true : false; }
// vector
template<class T> istream& operator >> (istream &is , vector<T> &v) { for(T &a : v) is >> a; return is; }
template<class T> ostream& operator << (ostream &os , const vector<T> &v) { for(const T &t : v) os << "\t" << t; return os << endl; }
// pair
template<class T, class U> ostream& operator << (ostream &os , const pair<T, U> &v) { return os << "<" << v.first << ", " << v.second << ">"; }

const int INF = 1 << 30;
const ll INFL = 1LL << 60;


vector<int> go(int N, ll K, const vector<ll> A, int start) {
    map<ll, int> s;
    vector<int> stk;
    for(ll i = start; i < N * K; i++) {
        ll a = A[i % N];
        if(s[a] > 0) {
            int sa = s[a];
            while(stk.size() >= sa) {
                s[stk.back()] = 0;
                stk.pop_back();
            }
        } else {
            stk.push_back(a);
            s[a] = stk.size();
        }
        // cerr << i << ": " << stk << endl;
    }
    return stk;
}

int main() {
    int N; ll K; cin >> N >> K;
    vector<ll> A(N); cin >> A;

    vector<int> next_index(N);
    map<ll, int> last;
    for(int i = N * 2 - 1; i >= 0; i--) {
        if(i < N) next_index[i] = last[A[i]];
        last[A[i%N]] = i % N;
    }
    //debug(next_index);
    ull left = K * N;
    map<int, int> visited;
    vector<ull> log;
    int now = 0;
    while(left) {
        if(visited.count(now)) {
            // cerr << "LOOP at " << now << endl;
           ull sum = 0;
            repeat(i, visited[(now-1+N)%N], log.size()) sum += log[i];
            // cerr << "sum = " << sum << " left = " << left << endl;
            left = left % sum;
            // now = (now + 1) % N;
            // break;
        }
        int next = next_index[now];
        ull diff = (now == next ? N : (next - now + N) % N) + 1;
        if(diff > left) break;
        left -= diff;
        // cerr << now << " -> " << (next + 1) % N << " (+" << diff << "), left " << left << endl;
        visited[now] = log.size();
        now = (next + 1) % N;
        log.push_back(diff);
    }
    // cerr << "go (left = " << left << " at " << now << endl;
    auto ans =  go(N, left / N + (left % N > 0), A, now);
    rep(i, ans.size()) cout << ans[i] << (i == ans.size() - 1 ? "\n" : " ");
    return 0;
}