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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;
typedef int64_t i64;

#define ALLOC(size,type) ((type*)calloc((size),sizeof(type)))

void run (void) {
  i32 n;
  i64 k;
  scanf ("%" SCNi32 "%" SCNi64, &n, &k);
  i32 *a = ALLOC (2 * n, i32);
  for (i32 i = 0; i < n; ++i) {
    scanf ("%" SCNi32, a + i);
    a[n + i] = a[i];
  }
  const i32 m = 200000;
  i64 *next = ALLOC (2 * n, i64);
  i64 *memo = ALLOC (m + 1, i64);
  for (i32 i = 2 * n - 1; i >= 0; --i) {
    next[i] = memo[a[i]] - i + 1;
    memo[a[i]] = i;
  }
  i64 **dp = ALLOC (65, i64 *);
  dp[0] = next;
  for (i32 i = 0; dp[i][0] < n * k; ++i) {
    dp[i + 1] = ALLOC (n, i64);
    for (i32 j = 0; j < n; ++j) {
      dp[i + 1][j] = dp[i][j] + dp[i][(j + dp[i][j]) % n];
    }
  }
  i64 now = 0;
  i32 d = 0;
  while (dp[d][now] < n * k) ++d;
  while (d >= 0) {
    if (now + dp[d][now % n] < n * k) {
      now += dp[d][now % n];
    } else {
      d--;
    }
  }
  uint8_t *elem = ALLOC (m + 1, uint8_t);
  i32 *ans = ALLOC (n, i32);
  i32 len = 0;
  for (i64 i = now; i < n * k; ++i) {
    i32 x = i % n;
    if (elem[a[x]]) {
      while (ans[len - 1] != a[x]) {
        elem[ans[--len]] = 0;
      }
      elem[ans[--len]] = 0;
    } else {
      ans[len++] = a[x];
      elem[a[x]] = 1;
    }
  }
  for (i32 i = 0; i < len; ++i) {
    printf ("%" PRIi32, ans[i]);
    putchar (i == len - 1 ? '\n' : ' ');
  }
}

int main (void) {
  run();
  return 0;
}