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;

#define debug(n) cerr << #n << ':' << n << endl;
#define dline    cerr << __LINE__ << endl;
#define rep(i,s,e) for(int i = (int)(s); i < (int)(e); ++i)
#define per(i,s,e) for(int i = (int)(s)-1; i >= (int)(e); --i)

using ll = long long;
template<class T, class U> using P = pair<T,U>;
template<class T> using Heap = priority_queue<T>;
template<class T> using heaP = priority_queue<T,vector<T>,greater<T>>;

template<class T>
bool ChangeMax(T&a,const T&b){
  if(a >= b) return false;
  a = b;    return true;
}

template<class T>
bool ChangeMin(T&a,const T&b){
  if(a <= b) return false;
  a = b;    return true;  
}

template<class T, size_t N, class U>
void Fill(T (&a)[N], const U&v){
    fill((U*)a,(U*)(a+N),v);
}

template<class T>
istream& operator >> (istream&is, vector<T>&v){
  for(auto&e:v)is >> e;
  return is;
}

int main(){
  int n; cin >> n;
  ll k; cin >> k;
  vector<int> v(n);
  cin >> v;
  vector<set<int>> sv(200001);
  rep(i,0,n) sv[v[i]].insert(i);
  int cnt = 1;
  {
    int p = n-1;
    while(1){
      int val = v[p];
      auto it = lower_bound(sv[val].begin(), sv[val].end(), p);
      if(it == sv[val].begin()){
	cnt++;
	it = sv[val].end();	
      }
      it--;      
      p = *it;
      if(p == 0) break;
      p--;      
    }
  }
  int t = k % cnt;
  cnt = 1;
  int p = 0;
  vector<int> ans; 
  while(1){

    int val = v[p];

    auto it = upper_bound(sv[val].begin(), sv[val].end(), p);
    if(it == sv[val].end()){
      if(cnt >= t){
	p++;
	ans.push_back(val);
	if(p>=n)break;
	continue;
      } else {
	cnt++;
	it = sv[val].begin();
      }
    }
    p = *it;
    p++;
    if(p>=n){
      if(cnt>=t){
	break;
      }
      cnt++;
      p=0;
    }    
  }
  rep(i,0,ans.size()){
    if(i)cout << ' ';
    cout << ans[i];
  }
  cout << endl;
  return 0;
}