This is an interactive problem.

There is an unknown integer x (1≤ x≤ n). You want to find x.

At first, you have a set of integers \{1, 2, …, n\}. You can perform the following operations no more than 10000 times:

  * A a: find how many numbers are multiples of a in the current set. 
  * B a: find how many numbers are multiples of a in this set, and then delete all multiples of a, but x will never be deleted (even if it is a multiple of a). In this operation, a must be greater than 1. 
  * C a: it means that you know that x=a. This operation can be only performed once. 



Remember that in the operation of type B a>1 must hold.

Write a program, that will find the value of x.

Input

The first line contains one integer n (1≤ n≤ 10^5). The remaining parts of the input will be given throughout the interaction process.

Interaction

In each round, your program needs to print a line containing one uppercase letter A, B or C and an integer a (1≤ a≤ n for operations A and C, 2≤ a≤ n for operation B). This line desribes operation you make.

If your operation has type C your program should terminate immediately.

Else your program should read one line containing a single integer, which is the answer to your operation.

After outputting each line, don't forget to flush the output. To do it use:

  * fflush(stdout) in C/C++; 
  * System.out.flush() in Java; 
  * sys.stdout.flush() in Python; 
  * flush(output) in Pascal; 
  * See the documentation for other languages. 



It is guaranteed, that the number x is fixed and won't change during the interaction process.

Hacks:

To make a hack, use such input format:

The only line should contain two integers n, x (1 ≤ x ≤ n ≤ 10^5).

Example

Input


10

2

4

0

Output



B 4

A 2

A 8

C 4

Note

Note that to make the sample more clear, we added extra empty lines. You shouldn't print any extra empty lines during the interaction process.

In the first test n=10 and x=4.

Initially the set is: \{1,2,3,4,5,6,7,8,9,10\}.

In the first operation, you ask how many numbers are multiples of 4 and delete them. The answer is 2 because there are two numbers divisible by 4: \{4,8\}. 8 will be deleted but 4 won't, because the number x will never be deleted. Now the set is \{1,2,3,4,5,6,7,9,10\}.

In the second operation, you ask how many numbers are multiples of 2. The answer is 4 because there are four numbers divisible by 2: \{2,4,6,10\}.

In the third operation, you ask how many numbers are multiples of 8. The answer is 0 because there isn't any number divisible by 8 in the current set.

In the fourth operation, you know that x=4, which is the right answer.
/*
Keep solving problems.
*/

import java.util.*;
import java.io.*;

public class Interactive {

  BufferedReader br;
  PrintWriter out;
  StringTokenizer st;
  boolean eof;
  final long MOD = 1000L * 1000L * 1000L + 7;
  int[] dx = {0, -1, 0, 1};
  int[] dy = {1, 0, -1, 0};

  void solve() {
    int n = nextInt();
    // seive algorithm
    boolean[] isPrime = new boolean[1 + n];
    Arrays.fill(isPrime, true);
    List<Integer> primes = new ArrayList<>();
    for (int i = 2; i <= n; i++) {
      if (isPrime[i]) {
        primes.add(i);
        for (int j = 2 * i; j <= n; j += i) {
          isPrime[j] = false;
        }
      }
    }

    List<List<Integer>> buckets = new ArrayList<>();
    int sz = 100;
    for (int i = 0; i < primes.size(); i += sz) {
      List<Integer> bucket = new ArrayList<>();
      for (int j = 0; j < sz; j++) {
        if (i + j < primes.size()) {
          bucket.add(primes.get(i + j));
        }
      }
      buckets.add(bucket);
    }

    List<Integer> factor = new ArrayList<>();
    int alive = n;
    boolean[] isAlive = new boolean[1 + n];
    Arrays.fill(isAlive, true);
    int smallest = -1;
    for (List<Integer> bucket : buckets) {
      for (int prime : bucket) {
        outln("B " + prime);
        int cntB = nextInt();
        int div = 0;
        for (int i = prime; i <= n; i += prime) {
          if (isAlive[i]) {
            isAlive[i] = false;
            alive--;
            div++;
          }
        }

        if (cntB != div) {
          factor.add(prime);
        }
      }

      outln("A " + 1);
      int cnt = nextInt();

      if (alive != cnt && smallest == -1) {
        // Find the smallest one.
        for (int prime : bucket) {
          outln("A " + prime);
          int cntA = nextInt();
          if (cntA > 0) {
            smallest = prime;
            break;
          }
        }
      }
    }
    if (smallest != -1) {
      factor.add(smallest);
    }

    long res = 1;
    for (int p : factor) {
      if (p < 1000) {
        int p2 = p * p;
        while (p2 <= n) {
          outln("A " + p2);
          int cntA = nextInt();
          if (cntA == 0) {
            break;
          }
          p2 = p2 * p;
        }

        p2 = p2 / p;
        res = res * p2;
      } else {
        res = res * p;
      }
    }

    outln("C " + res);
  }

  void shuffle(int[] a) {
    int n = a.length;
    for(int i = 0; i < n; i++) {
      int r = i + (int) (Math.random() * (n - i));
      int tmp = a[i];
      a[i] = a[r];
      a[r] = tmp;
    }
  }

  long gcd(long a, long b) {
    while(a != 0 && b != 0) {
      long c = b;
      b = a % b;
      a = c;
    }
    return a + b;
  }

  private void outln(Object o) {
    out.println(o);
    out.flush();
  }
  private void out(Object o) {
    out.print(o);
    out.flush();
  }
  public Interactive() throws IOException {
    br = new BufferedReader(new InputStreamReader(System.in));
    out = new PrintWriter(System.out);
    solve();
    out.close();
  }
  public static void main(String[] args) throws IOException {
    new Interactive();
  }

  public long[] nextLongArr(int n) {
    long[] res = new long[n];
    for(int i = 0; i < n; i++)
      res[i] = nextLong();
    return res;
  }
  public int[] nextIntArr(int n) {
    int[] res = new int[n];
    for(int i = 0; i < n; i++)
      res[i] = nextInt();
    return res;
  }
  public String nextToken() {
      while (st == null || !st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (Exception e) {
          eof = true;
          return null;
        }
      }
      return st.nextToken();
  }
  public String nextString() {
    try {
      return br.readLine();
    } catch (IOException e) {
      eof = true;
      return null;
    }
  }
  public int nextInt() {
      return Integer.parseInt(nextToken());
  }
  public long nextLong() {
      return Long.parseLong(nextToken());
  }
  public double nextDouble() {
      return Double.parseDouble(nextToken());
  }
}


