Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n;
bool isp[100005] = {0};
vector<long long> pr;
int main(void) {
  scanf("%lld", &n);
  for (long long i = 2; i <= n; i++) {
    if (isp[i] == 0) {
      pr.push_back(i);
      for (long long j = i * i; j <= n; j += i) isp[j] = 1;
    }
  }
  for (long long i = 1; i <= n; i++) isp[i] = 1;
  long long x;
  if (n <= 10) {
    for (long long i = n; i >= 2; i--) {
      printf("B %lld\n", i);
      fflush(stdout);
      scanf("%lld", &x);
      printf("A %lld\n", i);
      fflush(stdout);
      scanf("%lld", &x);
      if (x == 1) {
        printf("C %lld\n", i);
        fflush(stdout);
        return 0;
      }
    }
    printf("C 1\n");
    fflush(stdout);
    return 0;
  }
  long long siz = 90;
  long long now = n;
  long long ans = 1;
  for (long long i = 1; i * siz < pr.size(); i++) {
    long long upp = min((long long)pr.size(), (i + 1) * siz);
    for (long long j = i * siz; j < upp; j++) {
      printf("B %lld\n", pr[j]);
      fflush(stdout);
      scanf("%lld", &x);
      for (long long k = pr[j]; k <= n; k += pr[j]) {
        if (isp[k] == 1) isp[k] = 0, now--;
      }
    }
    printf("A 1\n");
    fflush(stdout);
    scanf("%lld", &x);
    if (x != now) {
      for (long long j = i * siz; j < upp; j++) {
        printf("A %lld\n", pr[j]);
        fflush(stdout);
        scanf("%lld", &x);
        long long no1 = pr[j];
        while (x != 0) {
          ans *= pr[j];
          no1 *= pr[j];
          if (no1 > n) break;
          ;
          printf("A %lld\n", no1);
          fflush(stdout);
          scanf("%lld", &x);
          if (x == 0) break;
        }
      }
      break;
    }
  }
  for (long long j = 0; j < min((long long)pr.size(), siz); j++) {
    printf("B %lld\n", pr[j]);
    fflush(stdout);
    scanf("%lld", &x);
    printf("A %lld\n", pr[j]);
    fflush(stdout);
    scanf("%lld", &x);
    long long no1 = pr[j];
    while (x != 0) {
      ans *= pr[j];
      no1 *= pr[j];
      if (no1 > n) break;
      ;
      printf("A %lld\n", no1);
      fflush(stdout);
      scanf("%lld", &x);
      if (x == 0) break;
    }
  }
  printf("C %lld\n", ans);
  fflush(stdout);
}