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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int num = sc.nextInt();
		long countMax = sc.nextLong();
		long array[] = new long[num];
		long sorted[][] = new long[num][2];
		for(int i = 0; i < num; i ++) {
			array[i] = sc.nextLong();
			sorted[i][0] = array[i];
			sorted[i][1] = i;
		}
		Arrays.sort(sorted, (a, b) -> Long.compare(a[0], b[0]));
		int next[] = new int[num];
		int prv = (int)sorted[0][1];
		for(int i = 1; i < num; i ++) {
			if(sorted[i][0] == sorted[i - 1][0]) {
				next[(int)sorted[i - 1][1]] = (int)sorted[i][1];
			}else {
				next[(int)sorted[i - 1][1]] = prv;
				prv = (int)sorted[i][1];
			}
		}
		next[(int)sorted[num - 1][1]] = prv;

		int count = 1;
		int tmpIndex = 0;
		int lastIndex[] = new int[num];
		while(tmpIndex != num) {
			if(next[tmpIndex] <= tmpIndex) {
				lastIndex[count - 1] = tmpIndex;
				count ++;
			}
			tmpIndex = next[tmpIndex] + 1;
		}
		countMax = (countMax - 1) % count;
		if(countMax == count - 1) {
			System.out.println(); return;
		}

		StringBuilder ans = new StringBuilder();
		tmpIndex = lastIndex[(int)countMax];
		while(tmpIndex < num) {
			if(next[tmpIndex] <= tmpIndex) {
				ans.append(array[tmpIndex]+" ");
				tmpIndex ++;
			}else {
				tmpIndex = next[tmpIndex] + 1;
			}
		}
		System.out.println(ans);
	}
}