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: Time Limit Exceeded)

#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}

use std::collections::HashSet;
fn main() {
    let v = read_vec::<usize>();
    let n = v[0];
    let k = v[1];

    let a = read_vec::<u64>();


    let mut i:usize = 1;
    let mut p: u64 = a[0];
    let mut blocksize: usize = 0;

    if a.iter().collect::<HashSet<&u64>>().len() == n {
        blocksize = n * (n + 1);
    } else {
        loop {
            let ia = i % n;
            if a[ia] == p {
                if ia == n - 1 {
                    blocksize = i + 1;
                    break;
                } else {
                    p = a[ia + 1];
                    i += 2;
                }
            } else {
                i += 1;
            }
        }
    }

    let rest = n * k % blocksize;

    let mut s = vec![];
    for i in 0..rest {
        let aa = a[i % n];
        if s.contains(&aa) {
            while s.pop().unwrap() != aa {}
        }
        else {
            s.push(a[i % n])
        }
    }
    let ans = s.into_iter().map(|i| i.to_string()).collect::<Vec<String>>().join(" ");
    println!("{}", ans);
}