Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = double;
const ll MODBASE = 1000000007LL;
const int MAXN = 100010;
const int MAXM = 1000;
const int MAXK = 16;
const int MAXQ = 200010;
int nt[MAXN], visited[MAXN];
vector<int> primes;
bool foundSmallest = false;
void sieveBase(int n) {
  for (int i = (2); i <= (n); i++) {
    if (nt[i] == 0) {
      nt[i] = i;
      primes.push_back(i);
      for (ll j = (ll)i * i; j <= n; j += i) nt[j] = i;
    }
  }
}
int ask(char ch, int x) {
  cout << ch << ' ' << x << endl;
  int g;
  cin >> g;
  return g;
}
void answer(int x) { cout << "C " << x << endl; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int n;
  cin >> n;
  sieveBase(n);
  int k = int(sqrt(int((primes).size())));
  int res = 1, sum = n;
  memset(visited, 0, sizeof(visited));
  for (int i = (0); i <= (int((primes).size()) - 1); i++) {
    int x = ask('B', primes[i]);
    int num = 0;
    for (int j = primes[i]; j <= n; j += primes[i])
      if (!visited[j]) {
        num++;
        sum--;
        visited[j] = 1;
      }
    if (x != num) {
      for (ll w = primes[i]; w <= n; w *= primes[i]) {
        int y = ask('A', w);
        if (y)
          res *= primes[i];
        else
          break;
      }
    }
    if ((i % k == k - 1 || i == int((primes).size()) - 1) && !foundSmallest) {
      int y = ask('A', 1);
      if (y != sum) {
        for (int j = i - k + 1; j <= i; j++) {
          for (ll w = primes[j]; w <= n; w *= primes[j]) {
            int z = ask('A', w);
            if (z) {
              res *= primes[j];
              foundSmallest = true;
            } else
              break;
          }
          if (foundSmallest) break;
        }
      }
    }
  }
  answer(res);
  return 0;
}