description: <image>
  
  This is an interactive task
  
  William has a certain sequence of integers a_1, a_2, ..., a_n in his mind, but due to security concerns, he does not want to reveal it to you completely. William is ready to respond to no more than 2 ⋅ n of the following questions:
  
    * What is the result of a [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of two items with indices i and j (i ≠ j) 
    * What is the result of a [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of two items with indices i and j (i ≠ j) 
  
  
  
  You can ask William these questions and you need to find the k-th smallest number of the sequence.
  
  Formally the k-th smallest number is equal to the number at the k-th place in a 1-indexed array sorted in non-decreasing order. For example in array [5, 3, 3, 10, 1] 4th smallest number is equal to 5, and 2nd and 3rd are 3.
  
  Input
  
  It is guaranteed that for each element in a sequence the condition 0 ≤ a_i ≤ 10^9 is satisfied.
  
  Interaction
  
  In the first line you will be given two integers n and k (3 ≤ n ≤ 10^4, 1 ≤ k ≤ n), which are the number of items in the sequence a and the number k.
  
  After that, you can ask no more than 2 ⋅ n questions (not including the "finish" operation).
  
  Each line of your output may be of one of the following types: 
  
    * "or i j" (1 ≤ i, j ≤ n, i ≠ j), where i and j are indices of items for which you want to calculate the bitwise OR. 
    * "and i j" (1 ≤ i, j ≤ n, i ≠ j), where i and j are indices of items for which you want to calculate the bitwise AND. 
    * "finish res", where res is the kth smallest number in the sequence. After outputting this line the program execution must conclude. 
  
  
  
  In response to the first two types of queries, you will get an integer x, the result of the operation for the numbers you have selected.
  
  After outputting a line do not forget to output a new line character and flush the output buffer. Otherwise you will get the "Idleness limit exceeded". To flush the buffer use:
  
    * fflush(stdout) in C++ 
    * System.out.flush() in Java 
    * stdout.flush() in Python 
    * flush(output) in Pascal 
    * for other languages refer to documentation 
  
  
  
  If you perform an incorrect query the response will be -1. After receiving response -1 you must immediately halt your program in order to receive an "Incorrect answer" verdict.
  
  Hacking
  
  To perform a hack you will need to use the following format:
  
  The first line must contain two integers n and k (3 ≤ n ≤ 10^4, 1 ≤ k ≤ n), which are the number of items in the sequence a and the number k.
  
  The second line must contain n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), the sequence a.
  
  Example
  
  Input
  
  
  7 6
  
  2
  
  7
  
  Output
  
  
  and 2 5
  
  or 5 6
  
  finish 5
  
  Note
  
  In the example, the hidden sequence is [1, 6, 4, 2, 3, 5, 4].
  
  Below is the interaction in the example.
  
  Query (contestant's program)| Response (interactor)| Notes  
  ---|---|---  
  and 2 5| 2| a_2=6, a_5=3. Interactor returns bitwise AND of the given numbers.  
  or 5 6| 7| a_5=3, a_6=5. Interactor returns bitwise OR of the given numbers.  
  finish 5| | 5 is the correct answer. Note that you must find the value and not the index of the kth smallest number.
solution: /*
          "Everything in the universe is balanced. Every disappointment
                  you face in life will be balanced by something good for you!
                          Keep going, never give up."
  */
  
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Codechef {
      public static void main(String[] args) throws java.lang.Exception {
          out = new PrintWriter(new BufferedOutputStream(System.out));
          sc = new FastReader();
  
          int test = 1;
          for (int t = 0; t < test; t++) {
              solve();
          }
          out.close();
      }
  
      private static void solve() {
          int n = sc.nextInt();
          int k = sc.nextInt();
          long x = getSum(1, 2);
          long y = getSum(2, 3);
          long z = getSum(1, 3);
  
          List<Long> list = new ArrayList<>();
          long second = (x + y - z) / 2;
          long first = x - second;
          long third = y - second;
          list.add(first);
          list.add(second);
          list.add(third);
          for (int i = 4; i <= n; i++) {
              long sum = getSum(1, i);
              list.add(sum - first);
          }
          Collections.sort(list);
          out.println("finish " + list.get(k - 1));
      }
  
      private static long getSum(int a, int b) {
          int and = ask(a, b, true);
          int or = ask(a, b, false);
          return and + or;
      }
  
      private static int ask(int a, int b, boolean and) {
          if (and) {
              out.print("and ");
          }else {
              out.print("or ");
          }
          out.println(a + " " + b);
          out.flush();
          int val = sc.nextInt();
          return val;
      }
  
      public static FastReader sc;
      public static PrintWriter out;
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              br = new BufferedReader(new
                      InputStreamReader(System.in));
          }
  
          String next()
          {
              while (st == null || !st.hasMoreElements())
              {
                  try
                  {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException  e)
                  {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt()
          {
              return Integer.parseInt(next());
          }
  
          long nextLong()
          {
              return Long.parseLong(next());
          }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine()
          {
              String str = "";
              try
              {
                  str = br.readLine();
              }
              catch (IOException e)
              {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }