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;
int dfs(int k, vector<int> &v, int n, int s, long long p, int t) {
  int r = 0;
  if (s % 2) {
    r = n / p;
  } else {
    r = -n / p;
  }
  for (int i = t; i < k && p * v[i] <= n; ++i) {
    r += dfs(k, v, n, s + 1, p * v[i], i + 1);
  }
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<bool> vj(n + 1, true);
  vector<bool> a(n + 1, true);
  for (int i = 2; i <= n; ++i)
    if (a[i]) {
      for (int j = 2; i * j <= n; ++j) a[i * j] = false;
    }
  vector<int> v;
  for (int i = 2; i <= n; ++i)
    if (a[i]) {
      v.push_back(i);
    }
  int m = v.size();
  vector<bool> isp(n + 1, false);
  int first = true;
  long long ans = 1;
  int t = 0;
  int sum = 0;
  for (int i = 0; i < m; ++i) {
    cout << "B " << v[i] << endl;
    int k;
    cin >> k;
    int cur = dfs(i, v, n, 1, v[i] * 1ll, 0);
    sum += cur;
    if (k != cur) {
      isp[v[i]] = true;
      ans *= v[i];
    }
    ++t;
    if ((t * t >= m || i == m - 1) && first) {
      cout << "A 1" << endl;
      cin >> k;
      if (n - k != sum) {
        for (int j = i - t + 1; j <= i; ++j) {
          cout << "A " << v[j] << endl;
          cin >> k;
          if (k) {
            isp[v[j]] = true;
            ans *= v[j];
            break;
          }
        }
        first = false;
      }
      t = 0;
    }
  }
  if (first) {
    cout << "C 1" << endl;
    return 0;
  }
  for (int i = 2; i <= n; ++i)
    if (isp[i]) {
      while (ans * i <= n) {
        cout << "A " << ans * i << endl;
        int k;
        cin >> k;
        if (k)
          ans *= i;
        else
          break;
      }
    }
  cout << "C " << ans << endl;
  return 0;
}
