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)


//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

// ここまで

fn run() {
    input! {
        n: usize,
        k: usize,
        a: [usize; n],
    }
    let mut b = a.clone();
    let mut c = a.clone();
    b.append(&mut c);
    let m = 200000 + 1;
    let mut next = vec![2 * n + 1; m];
    for i in (n..(2 * n)).rev() {
        next[b[i]] = i;
    }
    let mut dp = vec![vec![]; 60];
    dp[0].resize(n, 0);
    for i in (0..n).rev() {
        dp[0][i] = next[b[i]] - i + 1;
        next[b[i]] = i;
    }
    let mut depth = 0;
    while dp[depth][0] < n * k {
        depth += 1;
        dp[depth].resize(n, 0);
        for i in 0..n {
            let x = dp[depth - 1][i];
            let y = dp[depth - 1][(i + x) % n];
            dp[depth][i] = x + y;
        }
    }
    let mut step = 0;
    loop {
        if step + dp[depth][step % n] < n * k {
            step += dp[depth][step % n];
        } else if depth == 0 {
            break;
        } else {
            depth -= 1;
        }
    }
    let mut elem = vec![false; m];
    let mut ans = vec![];
    while step < n * k {
        let x = step % n;
        let v = a[x];
        step += 1;
        if elem[v] {
            while let Some(z) = ans.pop() {
                elem[z] = false;
                if v == z {
                    break;
                }
            }
        } else {
            elem[v] = true;
            ans.push(v);
        }
    }
    let mut out = String::new();
    for v in ans {
        out.push_str(&v.to_string());
        out.push(' ');
    }
    out.pop();
    println!("{}", out);
}

fn main() {
    run();
}