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 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
int main()
{
	long long int n, k;
	scanf("%lld %lld", &n, &k);
	long long int i, j;
	int a[200005];
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	long long int count;
	int b[200005];
	int c[200005];
	for (i = 0; i < 200005; i++)
		c[i] = 0;
	long long int l = 0;
	for (count = 0; count < k; count++)
	{
		for (i = 0; i < n; i++)
		{
			if (c[a[i]] == 0)
			{
				c[a[i]] = 1;
				b[l] = a[i];
				l++;
			}
			else
			{
				while (b[l - 1] != a[i])
				{
					l--;
					c[b[l]] = 0;
				}
				l--;
				c[b[l]] = 0;
			}
		}
		if (l == 0)
			break;
	}
	if (count < k)
	{
		k %= count + 1;
		l = 0;
		for (i = 0; i < 200005; i++)
			c[i] = 0;
		for (j = 0; j < k; j++)
		{
			for (i = 0; i < n; i++)
			{
				if (c[a[i]] == 0)
				{
					c[a[i]] = 1;
					b[l] = a[i];
					l++;
				}
				else
				{
					while (b[l - 1] != a[i])
					{
						l--;
						c[b[l]] = 0;
					}
					l--;
					c[b[l]] = 0;
				}
			}
		}
		if (l > 0)
		{
			for (i = 0; i < l - 1; i++)
				printf("%d ", b[i]);
			printf("%d\n", b[l - 1]);
		}
		else
			printf("\n");
	}
	else
	{
		if (l > 0)
		{
			for (i = 0; i < l - 1; i++)
				printf("%d ", b[i]);
			printf("%d\n", b[l - 1]);
		}
		else
			printf("\n");
	}
	return 0;
}