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;
typedef long long ll;
const ll INF = 1e18;
const int BIG = 2e9;
const int maxN = 2e5 + 55;

ll k, m;
ll a[2 * maxN];
int n;
vector<int> pos[maxN];
vector<pair<int, ll> > jump;

pair<int, ll> nxtPos(int i){
    int x = lower_bound(pos[a[i]].begin(), pos[a[i]].end(), i) - pos[a[i]].begin();
    int j = pos[a[i]][x + 1];
    ll len = j - pos[a[i]][x] + 1;
    return make_pair((i + len) % n, len);
}

void preCal(){
    jump.push_back(nxtPos(0));
    while(true){
        int pos = jump.back().first;
        if(pos == 0){
            break;
        }
        jump.push_back(nxtPos(pos));
    }
}

deque<int> ans;
bool mrk[maxN];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> k;

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

    for(int i = 0; i < n + n; i++){
        pos[a[i]].push_back(i);
    }

    preCal();

//    for(int i = 0, L = (int)jump.size(); i < L; i++){
//        //cout << jump[i].first << ' ' << jump[i].second << '\n';
//        m += jump[i].second;
//    }
//
//    ll rem = (((ll)n % m) * (k % m)) % m;
//
//    int from = 0;
//    ll psum = 0, step = 0;
//
//    for(int i = 0, L = (int)jump.size(); i < L; i++){
//        psum += jump[i].second;
//        if(psum <= rem){
//            from = jump[i].first;
//            step = rem - psum;
//        }
//    }
//
//    for(ll i = 1; i <= step; i++, from = (from + 1) % n){
//        if(mrk[a[from]]){
//            while(!ans.empty() && ans.back() != a[from]){
//                ans.pop_back();
//            }
//            ans.pop_back();
//            mrk[a[from]] = false;
//        } else {
//            ans.push_back(a[from]);
//            mrk[a[from]] = true;
//        }
//    }
//
//    while(!ans.empty()){
//        cout << ans.front() << ' ';
//        ans.pop_front();
//    }

    return 0;
}