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 long long N = 100001, M = 1005;
long long t, n, m;
vector<long long> prime;
bool is_comp[N];
void seive() {
  is_comp[0] = 1;
  is_comp[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!is_comp[i]) prime.push_back(i);
    long long sz = prime.size();
    for (int j = 0; j < sz && i * prime[j] <= n; j++) {
      is_comp[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
long long pw[25];
bool vis[N];
void del(int u) {
  for (int i = u; i <= n; i += u) {
    if (vis[i] == 0) m--;
    vis[i] = 1;
  }
  return;
}
int main() {
  cin >> n;
  seive();
  m = n;
  if (n == 1) {
    cout << "C 1"
         << "\n";
    cout << flush;
    return 0;
  }
  long long cur = 1, cnt, sz = (long long)prime.size(), ss = sqrt(n);
  long long k = 0, sq = sqrt(sz);
  for (int i = sz - 1; i >= 0 && prime[i] > ss; i--) {
    k++;
    del(prime[i]);
    cout << "B " << prime[i] << "\n";
    cout << flush;
    cin >> cnt;
    if (k == sq || (i == 0 || prime[i - 1] < ss)) {
      cout << "A 1"
           << "\n";
      cout << flush;
      cin >> cnt;
      if (m != cnt) {
        for (int j = 0; j < k; j++) {
          cout << "A " << prime[i + j] << "\n";
          cout << flush;
          cin >> cnt;
          if (cnt) {
            cur *= prime[i + j];
          }
        }
        break;
      }
      k = 0;
    }
  }
  for (auto p : prime) {
    if (p > n / cur || p > ss) break;
    cout << "B " << p << "\n";
    cout << flush;
    cin >> cnt;
    cout << "B " << p << "\n";
    cout << flush;
    cin >> cnt;
    if (cnt) {
      long long high = 0, low = 0, ans = 0;
      pw[0] = p;
      for (int i = 1; i < 20; i++) {
        if (pw[i - 1] * p > n / cur) break;
        pw[i] = pw[i - 1] * p;
        high = i;
      }
      while (high >= low) {
        long long mid = (high + low) / 2;
        cout << "B " << pw[mid] << "\n";
        cout << flush;
        cin >> cnt;
        if (cnt) {
          ans = mid;
          low = mid + 1;
        } else
          high = mid - 1;
      }
      cur *= pw[ans];
    }
  }
  cout << "C " << cur << "\n";
  cout << flush;
  return 0;
}
