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: Runtime Error)

import java.util.*

fun mul(A: Array<Int>, B: Array<Int>): Array<Int> {
    val C = Array(A.size) { it }

    for (i in A.indices) {
        C[i] = B[A[i]]
    }

    return C
}

fun pow(A: Array<Int>, k: Long): Array<Int> {
    var R = Array(A.size) { it }
    var B = A
    var l = k

    while (l > 0) {
        if (l and 1 != 0L) {
            R = mul(R, B)
        }
        B = mul(B, B)
        l /= 2
    }

    return R
}

fun main(args: Array<String>) {
    val (N, K) = readLine()!!.split(" ").map{it.toLong()}
    val A = readLine()!!.split(" ").map{it.toInt()}
    val ts = Array(100010) { TreeSet<Int>() }
    val mp = Array(A.size + 1) { it }

    A.forEachIndexed { index, v -> ts[v].add(index) }

    mp[A.size] = 0
    for (i in A.indices.reversed()) {
        if (i == A.size - 1) {
            mp[i] = ts[A[i]].first() + 1

            continue
        }

        val a = ts[A[i]].higher(i)
        if (a != null) {
            if (a >= mp.size - 1) {
                mp[i] = 0
            } else {
                mp[i] = mp[a + 1]
            }
        } else {
            mp[i] = ts[A[i]].first() + 1
        }
    }

    var B = pow(mp, K - 1)
    if (B[0] == B.size - 1) {
        B = pow(mp, K - 2)
    }

    val S = ArrayDeque<Int>()
    val st = TreeSet<Int>()

    for (i in B.first() % A.size until A.size) {
        if (st.contains(A[i])) {
            while (S.isNotEmpty()) {
                st.remove(S.last)
                if (S.pollLast() == A[i]) {
                    break
                }
            } } else {
            S.addLast(A[i])
            st.add(S.last)
        }
    }

    println(S.joinToString(" "))
}