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 + 10;
const int inf = 1e9;
const int mo = 998244353;
bool b[N];
int prime[N], primecnt = 0, vis[N], ncnt[N] = {0};
void eluer(int n) {
  for (long long i = 2; i <= n; i++) {
    if (!vis[i]) {
      prime[++primecnt] = i;
      for (long long j = i * i; j <= n; j += i) {
        vis[j] = 1;
      }
    }
  }
}
int main() {
  int n;
  cin >> n;
  eluer(n);
  for (auto i = (1); i <= (min(170, primecnt)); ++i) {
    cout << "B " << prime[i] << endl;
    cout.flush();
    int tmp;
    cin >> tmp;
  }
  int ans = 1;
  for (auto i = (1); i <= (min(170, primecnt)); ++i) {
    int now = prime[i];
    while (now <= n) {
      cout << "A " << now << endl;
      cout.flush();
      int x;
      cin >> x;
      if (x) {
        now *= prime[i];
        ans *= prime[i];
      } else
        break;
    }
  }
  if (primecnt > 170 && ans == 1) {
    int l = 171, r = primecnt, pdt = r - l + 1 + 1;
    while (l <= r) {
      int mid = l + r >> 1;
      for (auto i = (l); i <= (mid); ++i) {
        pdt--;
        cout << "B " << prime[i] << endl;
        cout.flush();
        int x;
        cin >> x;
      }
      cout << "A 1" << endl;
      cout.flush();
      int x;
      cin >> x;
      if (x == pdt) {
        l = mid + 1;
      } else {
        for (auto i = (l); i <= (mid); ++i) {
          cout << "A " << prime[i] << endl;
          cout.flush();
          int x;
          cin >> x;
          if (x) {
            ans *= prime[i];
            break;
          }
        }
        break;
      }
    }
  } else if (primecnt > 170) {
    for (auto i = (171); i <= (primecnt); ++i) {
      cout << "A " << prime[i] << endl;
      cout.flush();
      int x;
      cin >> x;
      if (x >= 2) {
        ans *= prime[i];
        break;
      }
    }
  }
  cout << "C " << ans << endl;
  cout.flush();
  return 0;
}
