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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author __shangu__
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BDoNotDuplicate solver = new BDoNotDuplicate();
        solver.solve(1, in, out);
        out.close();
    }

    static class BDoNotDuplicate {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt(), a[] = new int[n * 2], prev[] = new int[(int) (2e5 + 10)];
            long K = in.nextLong(), dp[][] = new long[60][n];
            for (int i = 0; i < n; ++i) {
                a[i] = in.nextInt();
                a[i + n] = a[i];
            }

            Arrays.fill(dp[0], n + 1);
            Arrays.fill(prev, -1);
            for (int i = n * 2 - 1; i >= 0; --i) {
                if (i < n) {
                    dp[0][i] = prev[a[i]] - i + 1;
                }
                prev[a[i]] = i;
            }
            for (int k = 1; (1L << k) < n * K; ++k) {
                for (int i = 0; i < n; ++i) {
                    dp[k][i] = Math.min(dp[k - 1][(int) ((dp[k - 1][i] + i) % n)] + dp[k - 1][i], n * K + 1);
                }
            }

            long pos = 0;
            long steps = 0;
            while (true) {
                int k = 0;
                while (steps + dp[k][(int) pos] < n * (K - 1)) ++k;
                if (k == 0) break;
                steps += dp[k - 1][(int) pos];
                pos += dp[k - 1][(int) pos];
                pos %= n;
            }
            pos += dp[0][(int) pos];
            pos %= n;
            int ans[] = new int[n], tot = 0;

            while (pos < n) {
                if (dp[0][(int) pos] + pos <= n) pos += dp[0][(int) pos];
                else {
                    ans[tot++] = a[(int) pos];
                    ++pos;
                }
            }

            for (int i = 0; i < tot; ++i) out.printf("%d ", ans[i]);
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}