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;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
int n;
int ett[100001];
vector<int> p;
set<int> st;
int query(char c, int n) {
  cout << c << ' ' << n << endl;
  int res;
  cin >> res;
  return res;
}
int main() {
  cin >> n;
  if (n == 1) {
    cout << "C 1" << endl;
    return 0;
  }
  for (int i = 2; i <= n; i++) {
    if (ett[i]) continue;
    p.push_back(i);
    for (int j = i + i; j <= n; j += i) ett[j] = 1;
  }
  int i = 0;
  for (; p[i] * p[i] <= n; i++) {
    query('B', p[i]);
    for (int j = p[i]; j <= n; j += p[i]) st.insert(j);
  }
  int cnt = n - st.size();
  int res = query('A', 1);
  if (res != cnt) {
    long long ans = 1;
    vector<int> vec;
    for (int i = 0; p[i] * p[i] <= n; i++) {
      int res = query('A', p[i]);
      if (res) vec.push_back(p[i]), ans *= p[i];
    }
    for (int x : vec) {
      while (ans * x <= n) {
        int res = query('A', ans * x);
        if (res == 0) break;
        ans *= x;
      }
    }
    for (; i < p.size(); i++) {
      if (ans * p[i] > n) break;
      int res = query('A', ans * p[i]);
      if (res) ans *= p[i];
    }
    cout << "C " << ans << endl;
    return 0;
  }
  int mn = i;
  for (; i < p.size(); i++) {
    query('B', p[i]);
    cnt--;
    if (i % 100 == 0 || i == p.size() - 1) {
      int res = query('A', 1);
      if (res > cnt) {
        for (int j = max(mn, i - 99); j <= i; j++) {
          int res = query('A', p[j]);
          if (res) {
            cout << "C " << p[j] << endl;
            return 0;
          }
        }
      }
    }
  }
  cout << "C 1" << endl;
}
