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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
fun main(args: Array<String>){
	val cin = Scanner(System.`in`)
    val n = cin.nextInt()
    val k = cin.nextLong()

    val a = 2*100000 + 1

    val jumpto = Array<Int>(n, { -1 })
    val now_tail = Array<Int>(a, { -1 })
    val head = Array<Int>(a, { -1 })
    val values = Array<Int>(n, { -1 })

    for (i in 0..n-1) {
        val t = cin.nextInt()
        values[i] = t
        if (now_tail[t] != -1) {
            jumpto[now_tail[t]] = i
        }
        now_tail[t] = i
        if (head[t] == -1) {
            head[t] = i
        }
    }

    for (i in 0..n-1) {
        if (jumpto[i] == -1) {
            jumpto[i] = head[values[i]]
        }
    }

    val round = Array<Int>(n, { -1 })
    val passed = Array<Boolean>(n, { false })

    var now = 0
    var now_round = 0
    while(!passed[now]) {
        passed[now] = true
        round[now] = now_round
        if (now >= jumpto[now]) {
            now_round += 1
        }
        now = jumpto[now] + 1
        if (now == n) {
            now_round += 1
            now = 0
        }
    }
    val cycle = now_round - round[now]
    val first = round[now] + 1
    val last = (k - first) % cycle

    var exceed = last.toInt()
    while(true) {
        if (exceed == 0 && now >= jumpto[now]) {
            break
        }
        if (now >= jumpto[now]) {
            exceed -= 1
        }
        now = jumpto[now] + 1
        if (now == n) {
            exceed -= 1
            now = 0
            if (exceed == -1) {
                break
            }
        }
    }

    val ans: MutableList<Int> = mutableListOf()
    while(n > now && exceed >= 0) {
        if (jumpto[now] > now) {
            now = jumpto[now] + 1
        } else {
            ans.add(values[now])
            now += 1
        }
    }

    for (i in 0..ans.size-1) {
        print(ans[i])
        print(" ")
    }
    println()
}