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 a[100500];
int b[100500];
int ans[100500];
vector<long long> prime;
int main() {
  int n;
  cin >> n;
  prime.reserve(10000);
  for (long long i = 2; i <= n; i++) {
    if (a[i] == 0) {
      a[i] = 1;
      prime.push_back(i);
      for (long long j = i * i; j <= n; j += i) {
        a[j] = 1;
      }
    }
  }
  reverse(prime.begin(), prime.end());
  int stage = 1;
  int cnt = 0;
  int x = 1;
  int total = n;
  vector<long long> printed;
  for (auto p : prime) {
    cnt++;
    cout << "B " << p << '\n';
    fflush(stdout);
    printed.push_back(p);
    int val;
    cin >> val;
    int exp = 0;
    for (int j = p; j <= n; j += p) {
      if (b[j] == 0) {
        b[j] = 1;
        exp++;
        total--;
      }
    }
    if (val != exp) {
      ans[p]++;
      long long ppp = p * p;
      while (true) {
        cout << "A " << ppp << '\n';
        fflush(stdout);
        int val3;
        cin >> val3;
        if (val3 != 0) {
          ans[p]++;
        } else {
          break;
        }
        ppp *= p;
        if (ppp > n) break;
      }
    }
    if ((cnt == 100 || p == 2) && stage == 1) {
      cout << "A 1" << '\n';
      fflush(stdout);
      int val2;
      cin >> val2;
      if (val2 != total) {
        stage = 2;
        for (auto pp : printed) {
          if (ans[pp] != 0) continue;
          long long ppp = pp;
          while (true) {
            cout << "A " << ppp << '\n';
            fflush(stdout);
            int val3;
            cin >> val3;
            if (val3 != 0) {
              ans[pp]++;
            } else {
              break;
            }
            ppp *= pp;
            if (ppp > n) break;
          }
        }
      }
      printed.clear();
      cnt = 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    while (ans[i]-- > 0) x *= i;
  }
  cout << "C " << x << '\n';
  fflush(stdout);
}
