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.
#include <bits/stdc++.h>
using namespace std;
int n, tot;
int pri[100005], num[100005];
int vis[100005], lef[100005], vis2[100005];
void init() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) {
      pri[++tot] = i, num[i] = tot, lef[i] = 1;
      for (int j = i + i; j <= n; j += i) {
        if (i >= min(320, n) + 1 && vis2[j] == 0) {
          vis2[j] = 1;
          lef[i]++;
        }
        vis[j] = 1;
      }
    }
  }
}
long long ans = 1;
void solvebig() {
  int cnt = 0, pre = 0;
  printf("A %d\n", 1);
  fflush(stdout);
  scanf("%d", &pre);
  for (int i = min(320, n) + 1; i <= n; i++)
    if (!vis[i]) {
      int _;
      printf("B %d\n", i);
      fflush(stdout);
      scanf("%d", &_);
      pre -= lef[i];
      cnt++;
      if (cnt % 50 == 0 || pri[num[i] + 1] == 0) {
        int gg = 0;
        printf("A %d\n", 1);
        fflush(stdout);
        scanf("%d", &gg);
        if (pre != gg) {
          for (int j = i; j >= 1; j--)
            if (!vis[j]) {
              int x;
              printf("A %d\n", j);
              fflush(stdout);
              scanf("%d", &x);
              if (x != 0) {
                ans *= j;
                break;
              }
            }
          return;
        }
      }
    }
}
void solvesmall() {
  for (int i = 2; i <= min(320, n); i++)
    if (!vis[i]) {
      int x;
      printf("B %d\n", i);
      fflush(stdout);
      scanf("%d", &x);
      printf("A %d\n", i);
      fflush(stdout);
      scanf("%d", &x);
      if (x != 0) {
        int temp = i;
        ans *= i;
        for (int j = 2; j <= 100000; j++) {
          temp = temp * i;
          if (temp > n) break;
          printf("A %d\n", temp);
          fflush(stdout);
          scanf("%d", &x);
          if (x == 0)
            break;
          else
            ans = ans * i;
        }
      }
    }
}
void output() {
  printf("C %lld\n", ans);
  fflush(stdout);
}
int main() {
  init();
  solvebig();
  solvesmall();
  output();
  return 0;
}
