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

#define NMAX 200000
int N;
long K;
int *A;
int s[NMAX];
int tp;
int *spat[NMAX];
int snum;
int previ[NMAX];
int cycle[NMAX];

int main(int argc, char *argv[]) {
	int sr;
	long i;
	int cy;
	int leap=0;
	void add(long);
	int add_pat(long);

	sr=scanf("%d %ld", &N, &K);
	A = (int *)malloc(sizeof(int)*N);
	for (i=0; i<N; i++) {
		sr=scanf("%d", A+i);
	}

	tp=0;
	snum=0;
	previ[0] = 0;
	for (i=0; i<N*K; i++) {
//if(i%1000000 == 0) fprintf(stderr, "%ld %d\n", i, tp);
		add(i);

		if (leap==0) {
			cy = add_pat(i);
			if (cy >= 0) {
//printf("[%d] %ld->", cy, i);
				for (i += ((N*K-i+1)/cy)*cy; i>= N*K;) i -= cy;
//printf("%ld\n", i);
				leap=1;
			}
		}
#if 0
printf("{%ld:%d} ", i, (int)(i%N));
for (int j=0; j<tp; j++) {
	printf("%d ", s[j]);
}
printf("\n");
#endif
	}

	if (tp>0) printf("%d", s[0]);
	for (i=1; i<tp; i++) {
		printf(" %d", s[i]);
	}
	printf("\n");

	return 0;

}
int X(long p) {
	return *(A+p%N);
}
void add(long n) {
	int a;
	int p;
	int i;
	int incl(int);

	a = X(n);
	p = incl(a);
	if (p<0) {
		s[tp] = a;
		tp++;
	} else {
		tp = p;
	}
}
int incl(int t) {
	int i;

	for (i=0; i<tp; i++) {
		if (s[i] == t) return i;
	}
	return -1;
}
int add_pat(long ai) {
	int i,j;

	for (i=0; i<snum; i++) {
// sと同じものがspat[i]の中にあるかをチェック
		for (j=0; j<tp && *(*(spat+i)+j) >= 0; j++) {
			if (s[j] != *(spat[i]+j)) break;
		}
		if (j<tp || *(spat[i]+j) != -1) continue;
		if (cycle[i] != ai%N) continue; //iの周期も一致するか
#if 0
printf("s=[");
for (j=0; j<tp; j++) printf("%d ", s[j]);
printf("](%d)\n", i);
#endif
		return ai-previ[i];
	}
	spat[snum] = (int *)malloc(sizeof(int)*tp+1);
	for (i=0; i<tp; i++) {
		*(spat[snum]+i) = s[i];
	}
	*(spat[snum]+tp) = -1;
	cycle[snum] = ai%N;
	previ[snum] = ai;
	snum++;
	return -1;
}