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 = 1e5 + 1;
int n, m, tot, ans = 1, tag[N], pri[N];
inline void input() { cin >> n, m = sqrt(n); }
inline void init() {
  for (int i = 2; i <= n; i++) {
    if (!tag[i]) pri[++tot] = i;
    for (int j = 1; j <= tot; j++) {
      if (i * pri[j] > n) break;
      tag[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
inline void del(int x, int out = 0) { cout << "B " << x << endl, cin >> out; }
inline int count(int x, int out = 0) {
  return cout << "A " << x << endl, cin >> out, out;
}
inline void solve(int x) {
  for (int p = 1; n / p / x; ans *= x)
    if (!count(p *= x)) return;
}
inline void work(int p = 0) {
  for (int i = 1; i <= tot; i++)
    if (pri[i] <= m) del(pri[i]);
  for (int i = 1; i <= tot; i++)
    if (pri[i] <= m) solve(pri[p = i]);
  for (m = sqrt(tot); pri[tot] > n / ans;) tot--;
  if (ans == 1) {
    for (m = sqrt(tot); pri[tot] > n / ans;) tot--;
    for (int l = p + 1, r, now = count(1); l <= tot; l = r + 1) {
      r = min(tot, l + m);
      for (int i = l; i <= r; i++) del(pri[i]), now--;
      if (count(1) > now) {
        for (int i = l; i <= r; i++)
          if (count(pri[i])) {
            ans *= pri[i];
            break;
          }
        break;
      }
    }
  } else
    for (int i = p + 1; i <= tot; i++)
      if (count(pri[i]) >= 2) ans *= pri[i];
  cout << "C " << ans << endl;
}
int main() {
  input();
  init();
  work();
  return 0;
}
