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;
using ll = long long;
ll INF = 1e9+7;
int main()
{
    ll n, k;
    cin >> n >> k;
    vector<ll> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    vector<ll> ar = a;
    reverse(ar.begin(), ar.end());
    vector<set<ll>> pos(2*1e5+10);
    for(int i = 0; i < n; i++)
    {
        pos[ar[i]].insert(i+1);
        if(pos[ar[i]].size() == 1)
        {
            pos[ar[i]].insert(i+1+n);
        }
    }

    vector<ll> start(n+1);
    int p = 1;
    ll v = ar[0];
    ll ans = 0;
    while(true)
    {
        ll next = *(pos[v].lower_bound(p+1));
        if(next == n)
        {
            start[ans] = next;   
            break;
        }
        else if(next == 2*n)
        {
            ans++;
            start[ans] = next - n;
            break;
        }
        else if(next > n)
        {
            p = next+1-n;
            ans++;
            start[ans] = p-1;
        }
        else
        {
            p = next+1;
            start[ans] = next;
        }
        v = ar[p-1];
    }
    ans++;
    ll count = k % ans;
    if(count == 0) return 0;
    ll s = start[ans-count];
    stack<ll> sta;
    set<ll> tmp;
    for(int i = n-s; i < n; i++)
    {
        if(tmp.find(a[i]) == tmp.end())
        {
            tmp.insert(a[i]);
            sta.push(a[i]);
        }
        else
        {
          while(true)
            {
                ll cand = sta.top();
                sta.pop();
                tmp.erase(cand);
                if(cand == a[i])
                {
                    break;
                }
            }
        }
    }
    stack<ll> ansq;
    ll size = sta.size();
    for(int i = 0; i < size; i++)
    {
        ll cand = sta.top();
        sta.pop();
        ansq.push(cand);
    }
    ll anss = ansq.size();
    for(int i = 0; i < anss; i++)
    {
        ll cand = ansq.top();
        ansq.pop();
        cout << cand << " ";
    }
}