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 B = 100;
int vis[N], p[N], tot = 0;
int used[N], all;
vector<int> por;
void sieve(int n) {
  vis[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) p[++tot] = i;
    for (int j = 1; j <= tot && p[j] * i <= n; j++) {
      vis[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
  return;
}
int main() {
  int n;
  cin >> n;
  all = n;
  int ans;
  sieve(n);
  int l = 0, r = B;
  for (int i = 1; i <= tot; i++) {
    cout << "B " << p[i] << endl;
    int cnt = 0;
    for (int j = p[i]; j <= n; j += p[i]) {
      if (used[j] == 0) used[j] = 1, cnt++, all--;
    }
    cin >> ans;
    if (ans > cnt) {
      por.push_back(p[i]);
    }
    if (i % B == 0) {
      cout << "A 1" << endl;
      cin >> ans;
      if (ans > all)
        r = min(r, i / B);
      else
        l = max(l, i / B);
    }
  }
  cout << "A 1" << endl;
  cin >> ans;
  if (ans == 1) return cout << "C 1" << endl, 0;
  for (int i = l * B + 1; i <= min(tot, r * B); i++) {
    cout << "A " << p[i] << endl;
    cin >> ans;
    if (ans >= 1) {
      por.push_back(p[i]);
      break;
    }
  }
  int siz = por.size();
  int cal = 1;
  for (int i = 0; i < siz; i++) {
    for (long long k = por[i]; k <= n; k *= por[i]) {
      cout << "A " << k << endl;
      cin >> ans;
      if (ans >= 1)
        cal *= por[i];
      else
        break;
    }
  }
  cout << "C " << cal << endl;
  return 0;
}
