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: Accepted)

use std::cmp::{max, min};
use std::io::prelude::*;

fn main() {
    let mut input_str = String::new();
    std::io::stdin().read_to_string(&mut input_str).unwrap();

    let input_parts = input_str.split_whitespace().collect::<Vec<_>>();
    let mut input_parts_it = input_parts.iter().cloned();

    let n: usize = input_parts_it.next().unwrap().parse().unwrap();
    let k: u64 = input_parts_it.next().unwrap().parse().unwrap();

    let mut a: Vec<u32> = (0..n).map(|_|  input_parts_it.next().unwrap().parse().unwrap()).collect();

    const NONE: usize = 0xffffffff;
    let mut group_end: Vec<usize> = vec![NONE; a.len()];

    let mut next_a_same_value_for_a: Vec<usize> = vec![NONE; a.len()];
    let mut last_a_for_aval: Vec<usize> = vec![NONE; 200001];

    for (i, &aval) in a.iter().enumerate().rev() {
        next_a_same_value_for_a[i] = last_a_for_aval[aval as usize];
        last_a_for_aval[aval as usize] = i;
    }

    for (i, &aval) in a.iter().enumerate() {
        // find the next `aval` elem with an identical value
        let mut k = next_a_same_value_for_a[i];
        if k == NONE {
            k = last_a_for_aval[aval as usize];
            debug_assert!(k != NONE);
            k += a.len();
        }
        k += 1;

        group_end[i] = k;
    }

    // check the cycle length
    let mut cycle_len = 0;
    {
        let mut i = 0;
        loop {
            let mut hop = group_end[i] - i;
            cycle_len += hop as u64;
            i = group_end[i];
            while i >= a.len() {
                i -= a.len();
            }
            if i == 0 {
                break;
            }
        }
    }

    // skip cycles
    let mut step = n as u64 * k;
    step = step %cycle_len;


    let mut state: Vec<u32> = Vec::with_capacity(a.len());
    let mut index: Vec<usize> = vec![NONE; 200001];

    {
        let mut i = 0;
        let mut remaining_steps = step;

        while remaining_steps >= (group_end[i] - i) as u64 {
            remaining_steps -= (group_end[i] - i) as u64;
            i = group_end[i];
            while i >= a.len() {
                i -= a.len();
            }
        }

        while remaining_steps > 0 {
            let aval = a[i];

            if index[aval as usize] == NONE {
                index[aval as usize] = state.len();
                state.push(aval);
            } else {
                while state.len() > index[aval as usize] {
                    let k = state.pop().unwrap();
                    index[k as usize] = NONE;
                }
            }

            i += 1;
            while i >= a.len() {
                i -= a.len();
            }
            remaining_steps -= 1;
        }
    }

    if state.is_empty() {
        println!("");
    } else {
        print!("{}", state[0]);
        for &x in state[1..].iter() {
            print!(" {}", x);
        }
    }
}