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)

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>


/* define */
#define Yes(X) puts ( (X) ? "Yes" : "No")
#define YES(X) puts ( (X) ? "YES" : "NO")
#define FOR(i, a, b) for ( int i = (a); i < (b) ; i++)
#define REP(i, n) for ( int i = 0; i < (n) ; i++)
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max3(a, b, c) ((max((a), (b)) > (c)) ? max((a), (b)) : (c))
#define min3(a, b, c) ((min((a), (b)) < (c)) ? min((a), (b)) : (c))

/* const */
const int MOD = 1000000007;

/* alias */
typedef long long ll;
typedef unsigned  u;

/* function */
int cmp(const void *a, const void *b) {
	return *(int*)a - *(int*)b;
}
int cmp2(const void *a, const void *b) {
	return *(int*)b - *(int*)a;
}
int gcd(int a, int b) {
	if (b != 0)return gcd(b, a % b);
	return a;
}
ll Lgcd(ll a, ll b) {
	if (b != 0) return Lgcd(b, a % b);
	return a;
}
int lcm(int a, int b) {
	return a / gcd(a, b) * b;
}
ll Llcm(ll a, ll b) {
	return a / Lgcd(a, b) * b;
}

/* main */

int n, a[200000], s[200000] = {}, c = 0;

int main() {
	ll k;
	scanf("%d%lld", &n, &k);
	REP(i, n)scanf("%d", &a[i]);
	REP(l, k) {
		for (int i = 0; i < n; i++) {
			int f = 1;
			REP(j, c) {
				if (a[i] == s[j]) {
					f = 0;
					c = j;
					break;
				}
			}
			if (f) {
				s[c] = a[i];
				c++;
			}
		}

		if (c == 0) {
			k %= (l + 1);
			l = -1;
		}
	}

	REP(i, c)printf("%d ", s[i]);
	return 0;
}