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 = 100005;
int lp[N + 1];
vector<long long> pr;
int main() {
  for (int i = 2; i <= N; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0; j < (int)pr.size() && pr[j] <= lp[i] && i * pr[j] <= N; ++j)
      lp[i * pr[j]] = pr[j];
  }
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (n == 1) {
    cout << "C 1" << endl;
    cout.flush();
    return 0;
  }
  while (pr.back() > n) pr.pop_back();
  vector<long long> factor;
  bool toggle[n + 1];
  memset(toggle, 0, sizeof(toggle));
  int i = 0;
  for (; i + 100 < pr.size(); i += 100) {
    for (int a = 0; a < 100; a++) {
      cout << "B " << pr[i + a] << endl;
      cout.flush();
      int res;
      cin >> res;
      for (int b = pr[i + a]; b <= n; b += pr[i + a]) toggle[b] = true;
    }
    int shouldbe = 0;
    for (int a = 1; a <= n; a++)
      if (!toggle[a]) shouldbe++;
    cout << "A 1" << endl;
    cout.flush();
    int actual;
    cin >> actual;
    if (actual != shouldbe) break;
  }
  for (; i < pr.size(); i++) {
    cout << "B " << pr[i] << endl;
    cout.flush();
    for (int a = pr[i]; a <= n; a += pr[i]) toggle[a] = true;
    int bef;
    cin >> bef;
    cout << "A " << pr[i] << endl;
    cout.flush();
    int ans;
    cin >> ans;
    if (ans != 0) {
      factor.push_back(pr[i]);
      i++;
      break;
    }
  }
  while (i < pr.size()) {
    cout << "B " << pr[i] << endl;
    cout.flush();
    int actual;
    cin >> actual;
    int shouldbe = 0;
    for (int a = pr[i]; a <= n; a += pr[i]) {
      if (!toggle[a]) shouldbe++, toggle[a] = true;
    }
    if (shouldbe != actual) factor.push_back(pr[i]);
    i++;
  }
  long long ans = 1;
  for (int i = 0; i < factor.size(); i++) {
    long long ite = 1;
    while (ite * factor[i] <= n) ite *= factor[i];
    for (; ite > 1; ite /= factor[i]) {
      cout << "A " << ite << endl;
      cout.flush();
      int res;
      cin >> res;
      if (res != 0) {
        ans *= ite;
        break;
      }
    }
  }
  cout << "C " << ans << endl;
}
