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: Time Limit Exceeded)

#include <cstdio>
#include <algorithm>
#include <stack>
#include <cstring>
#include <vector>
using namespace std;

typedef long long LL;

const int N = 2e5;

int n;
LL k;
int a[N+5];
bool cnt[N+5];

stack<int> Q;
stack<int> P;

vector<int> ans[N+5];

bool circle;
int tot;

bool Pan()
{
	if(ans[tot].size() != ans[1].size()) return false;
	for(register int i=ans[1].size()-1; i>=0; i--)
		if(ans[tot][i] != ans[1][i])
			return false;
	return true;
}

int main()
{
	
	scanf("%d %lld", &n, &k);
	
	int pp=0;
	for(register int i=1; i<=n; i++)
	{
		scanf("%d", &a[i]);
		if(!cnt[a[i]])
		{
			cnt[a[i]] = 1;
			pp++;
		}
	}
	if(pp == n)
	{
		k %= n+1;
		for(register int i=k; i<=n; i++)
			printf("%d ", a[i]);
		return 0;
	}
	
	memset(cnt, 0, sizeof cnt);
	for(register LL i=1; i<=k; i++)
	{
		for(register int j=1; j<=n; j++)
		{
			if(!cnt[a[j]])
			{
				Q.push(a[j]);
				cnt[a[j]] = 1;
				continue;
			}
			while(cnt[a[j]])
			{
				cnt[Q.top()] = 0;
				Q.pop();
			}
		}
		
		P = Q;
		tot++;
		
		while(!P.empty())
			ans[tot].push_back(P.top()), P.pop();
		
		if(i != 1 && Pan())
		{
			circle = 1;
			break;
		}
	}
	
	if(circle)
		k %= tot-1;
	
	for(register int i=ans[k].size()-1; i>=0; i--)
		printf("%d ", ans[k][i]);
	
	return 0;
}