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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
  "bufio"
  "os"
  "strconv"
  "strings"
  "fmt"
)

var reader = bufio.NewReaderSize(os.Stdin, 1000000)
var writer = bufio.NewWriter(os.Stdout)
func NextLine() string {
  var line, buffer []byte
  var isPrefix bool = true
  var err error
  for isPrefix {
    line, isPrefix, err = reader.ReadLine()
    if err != nil { panic(err) }
    buffer = append(buffer, line...)
  }
  return string(buffer)
}
func NextIntVec() []int {
  L := strings.Split(NextLine(), " ")
  N := make([]int, len(L))
  for i := range N {
    N[i], _ = strconv.Atoi(L[i])
  }
  return N
}
func Write(s interface{}) {
  fmt.Fprintln(writer, s)
}
func Output() {
  _ = writer.Flush()
}

func NextSearch(A *[]int, v int) int {
  n := len((*A))
  if (*A)[n - 1] == v {
    return 0
  }
  high := n
  low := 0
  mid := 0
  for ; high > low; {
    mid = (high + low) / 2
    if (*A)[mid] == v {
      break
    }
    if (*A)[mid] < v {
      low = mid
    } else {
      high = mid
    }
  }
  return mid + 1
}

func main() {
  NK := NextIntVec()
  N := NK[0]
  A := NextIntVec()
  D := make([][]int, 200001)
  for i := range D {
    D[i] = make([]int, 0)
  }
  for i, a := range A {
    D[a] = append(D[a], i)
  }
  k := 1
  v := -1
  for ; v < N - 1; {
    v += 1
    a := A[v]
    idx := NextSearch(&D[a], v)
    v = D[a][idx]
    if idx == 0 {
      k += 1
      if NK[1] < k { break }
    }
  }
  K := NK[1] % k
  v = 0
  for ; K > 0; v++ {
    a := A[v]
    idx := NextSearch(&D[a], v)
    if idx == 0 {
      K -= 1
      if K == 0 { break }
    }
    v = D[a][idx]
  }
  S := make([]int, 0)
  for ; v < N; v++ {
    a := A[v]
    idx := NextSearch(&D[a], v)
    if idx == 0 {
      S = append(S, a)
    } else {
      v = D[a][idx]
    }
  }
  ans := ""
  if len(S) > 0 {
    ans = strconv.Itoa(S[0])
    for i := 1; i < len(S); i++ {
      ans += " " + strconv.Itoa(S[i])
    }
  }
  Write(ans)
  Output()
}