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: Accepted)

#include <stdio.h>

typedef struct List {
	struct List *next;
	int v;
} list;

int main()
{
	int i, N, A[200001];
	long long K;
	scanf("%d %lld", &N, &K);
	for (i = 1; i <= N; i++) scanf("%d", &(A[i]));
	A[0] = 0;
	
	list *L[200001] = {}, d[200002], *p;
	d[0].v = 0;
	d[0].next = L[0];
	L[0] = &(d[0]);
	d[N+1].v = 0;
	d[N+1].next = NULL;
	for (i = N; i >= 1; i--) {
		d[i].v = i;
		d[i].next = L[A[i]];
		L[A[i]] = &(d[i]);
	}

	int j, k, l, move[200001];
	for (i = 0; i <= 200000; i++) move[i] = -1;
	for (i = 0, k = 0; move[i] == -1; i = move[i], k++) {
		for (p = &(d[L[i]->v + 1]); p->next != NULL; p = &(d[p->next->v + 1]));
		move[i] = A[p->v];
	}
	for (j = 0, l = 0; j != i; j = move[j], l++, k--);
	
	int s[200001];
	if (K >= l) K = (int)((K - l) % k + l);
	for (i = 0, j = 1; j < K; i = move[i], j++);
	j = L[i]->v + 1;
	k = 0;
	while (j <= N) {
		if (d[j].next == NULL) s[k++] = A[j++];
		else j = d[j].next->v + 1;
	}
	for (i = 0; i < k - 1; i++) printf("%d ", s[i]);
	if (k > 0) printf("%d\n", s[k-1]);
	fflush(stdout);
	return 0;
}