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 Rust (1.15.1) (judge verdict: Runtime Error)

macro_rules! get {
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t:ty),*) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            (
                $(iter.next().unwrap().parse::<$t>().unwrap(),)*
            )
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_|
                    get!($t)
        ).collect::<Vec<_>>()
    };
    ($($t:ty),*; $n:expr) => {
        (0..$n).map(|_|
                    get!($($t),*)
        ).collect::<Vec<_>>()
    };
    ($t:ty |) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
    ($t:ty | $n:expr) => {
        (0..$n).map(|_| get!($t |)).collect::<Vec<_>>()
    };
}

use std::collections::*;

fn main() {
    let (N, K) = get!(usize, u64);
    let As = get!(usize | );
    let mut s = VecDeque::new();
    let mut cycle = 0;
    let mut first_idxs = vec![None; 200000];
    while cycle == 0 || s.len() > 0 {
        cycle += 1;
        for (i, &A) in As.iter().enumerate() {
            if let Some(first_idx) = first_idxs[A] {
                loop {
                    let p = s.pop_back().unwrap();
                    first_idxs[p] = None;
                    if p == A {
                        break;
                    }
                }
            } else {
                s.push_back(A);
                first_idxs[A] = Some(i);
            }
        }
    }
    let mut first_idxs = vec![None; 200000];
    for t in 0..(K%cycle) {
        for (i, &A) in As.iter().enumerate() {
            if let Some(first_idx) = first_idxs[A] {
                loop {
                    let p = s.pop_back().unwrap();
                    first_idxs[p] = None;
                    if p == A {
                        break;
                    }
                }
            } else {
                s.push_back(A);
                first_idxs[A] = Some(i);
            }
        }
    }
    for s in s {
        print!("{} ", s);
    }
}