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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

constexpr int MAX = 200200;

int main() {
	int n;
	long long k;
	cin >> n >> k;
	vector<int> a(n);
	for (int &e : a) cin >> e;
	for (int i = 0; i < n; i++) {
		a.push_back(a[i]);
	}
	
	vector<int> pi(MAX, -1), next(n, -1);
	for (int i = 0; i < 2 * n; i++) {
		if (int j = pi[a[i]]; j != -1 and next[j] == -1) {
			next[j] = (i + 1) % n;
		}
		pi[a[i]] = i % n;
	}

	vector<bool> processed(n, false);
	for (int s = 0; s < n; s++) {
		if (processed[s]) continue;
		int i = s;
		vector<int> seen;
		for (; next[i] > i + 1; i = next[i]) {
			seen.push_back(i);
		}
		for (int j : seen) {
			processed[j] = true;
			next[j] = next[i];
		}
	}

	k--;
	int start = 0;
	for (int i = 0; i < 50; i++) {
		if ((k >> i) & 1) start = next[start];
		vector<int> buff(n);
		for (int j = 0; j < n; j++) {
			buff[j] = next[next[j]];
		}
		next.swap(buff);
	}
	
	vector<int> ans;
	vector<bool> pushed(MAX, false);
	for (int i = start; i < n; i++) {
		if (pushed[a[i]]) {
			while (ans.back() != a[i]) {
				pushed[ans.back()] = false;
				ans.pop_back();
			}
			pushed[a[i]] = false;
			ans.pop_back();
		} else {
			ans.push_back(a[i]);
			pushed[a[i]] = true;
		}
	}
	
	for (int e : ans) cout << e << ' ';
	
	return 0;
}