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;
const int N = 2e5 + 10;
vector<int> num;
bool st[N];
void init(int n) {
  for (int i = 2; i <= n; i++) {
    if (!st[i]) num.push_back(i);
    for (int j = 0; j < num.size() && num[j] <= n / i; j++) {
      st[i * num[j]] = true;
      if (i % num[j] == 0) break;
    }
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  init(n);
  int siz = num.size(), idx = -1;
  for (int i = 0; i < siz; i++) {
    if (num[i] <= sqrt(n))
      idx = i;
    else
      break;
  }
  int limit = 100;
  int cnt = n;
  int res = 1;
  for (int i = 1; i <= limit; i++) {
    int l = (i - 1) * limit, r = min(i * limit - 1, siz - 1);
    for (int j = l; j <= r; j++) {
      if (j <= idx) continue;
      printf("B %d\n", num[j]);
      fflush(stdout);
      int tt;
      scanf("%d", &tt);
      if (tt) cnt -= tt;
    }
    printf("A 1\n");
    fflush(stdout);
    int tt;
    scanf("%d", &tt);
    if (cnt != tt) {
      for (int j = l; j <= r; j++) {
        if (j <= idx) continue;
        printf("A %d\n", num[j]);
        fflush(stdout);
        int tt;
        scanf("%d", &tt);
        if (tt) {
          res = num[j];
          break;
        }
      }
      break;
    }
    if (r == siz - 1) break;
  }
  for (int i = 0; i <= idx; i++) {
    printf("B %d\n", num[i]);
    fflush(stdout);
    int tt;
    scanf("%d", &tt);
    printf("A %d\n", num[i]);
    fflush(stdout);
    scanf("%d", &tt);
    if (tt) {
      int x = num[i];
      while (x * num[i] <= n) {
        printf("B %d\n", x * num[i]);
        fflush(stdout);
        int ttt;
        scanf("%d", &ttt);
        if (ttt)
          x *= num[i];
        else
          break;
      }
      res = res * x;
    }
  }
  printf("C %d\n", res);
  fflush(stdout);
}
int main() {
  solve();
  return 0;
}
