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;


vector<int>solve2(const vector<int>&v,int from){
	vector<int>que;
	map<int,int>mp;
	for(int x=from;x<v.size();++x){
		if(mp.find(v[x])==mp.end()){
			que.emplace_back(v[x]);
			mp[v[x]]=que.size()-1;
		}else{
			while(true){
				int aback=que.back();
				
				mp.erase(aback);
				que.pop_back();
				if(aback==v[x])break;
			}
		}
	}
	return que;
}
vector<int> solve(int N,int K,vector<int>v,int start){
	
	vector<int>rs(N);

	{
		map<int,int>mp;
		for(int i=N-1;i>=0;--i){
			mp[v[i]]=N+i;
		}
		for(int i=N-1;i>=0;--i){
			rs[i]=mp[v[i]];
			mp[v[i]]=i;
		}
	}
	

	long long int ny=0,nx=start;
	map<int,long long int>mp;
	mp[start]=0;
	while(true){
		auto next_x=rs[nx];
		long long next_y=ny;
		//cout<<next_x<<' '<<next_y<<endl;
		if(next_x>=N){
			next_y++;
			next_x-=N;
		}
		next_x++;
		if(next_x>=N){
			next_y++;
			next_x-=N;
		}
		//cout<<K<<endl;
		if(next_y>=K){
			
			if(K-1==next_y){
				v.insert(v.end(),v.begin(),v.end());
			}
			return solve2(v,nx);
		}else{
			ny=next_y;
			nx=next_x;

			if(mp.find(nx)==mp.end()){
				mp[nx]=ny;
			}else{
				long long sa=-(mp[nx]-ny);
				long long int rest=K-ny;
				rest%=sa;
				if(rest==0)rest=sa;
				return solve(N,rest,v,nx);
			}
		}
	}
	assert(false);
	
}
int main() {
	int N;cin>>N;
	long long int K;cin>>K;
	vector<int>v;
	for(int i=0;i<N;++i){
		int a;cin>>a;
		v.push_back(a);
	}
	auto ans=solve(N,K,v,0);
	for(auto an:ans)cout<<an<<' ';
	cout<<endl;
	return 0;
}