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;
template <typename T>
void print(T a) {
  for (auto x : a) cout << x << ' ';
  cout << '\n';
}
long long ask_n() {
  cout << "A 1" << endl;
  long long ans;
  cin >> ans;
  return ans;
}
long long ask(long long a) {
  cout << "B " << a << endl;
  long long ans;
  cin >> ans;
  return ans;
}
void answer(long long a) { cout << "C " << a << endl; }
void solve() {
  vector<long long> primes;
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    bool prime = true;
    for (long long j = 2; j * j <= i; j++) {
      if (i % j == 0) prime = false;
    }
    if (prime) primes.push_back(i);
  }
  long long sqrt = 98;
  vector<bool> used(n + 1, false);
  long long free = n;
  vector<long long> factors;
  bool done = false;
  for (long long i = 0; i < (long long)primes.size(); i += sqrt) {
    for (long long j = 0; j < sqrt && (i + j) < (long long)primes.size(); j++) {
      long long cnt = ask(primes[i + j]);
      long long supposed = 0;
      for (long long k = primes[i + j]; k <= n; k += primes[i + j])
        if (!used[k]) {
          supposed++;
          free--;
          used[k] = true;
        }
      if (cnt != supposed) factors.push_back(primes[i + j]);
    }
    long long cnt = ask_n();
    if (cnt != free && !done) {
      done = true;
      for (long long j = 0; j < sqrt && (i + j) < (long long)primes.size();
           j++) {
        if (ask(primes[i + j])) {
          factors.push_back(primes[i + j]);
          break;
        }
      }
    }
  }
  long long ans = 1;
  for (long long x : factors) {
    for (long long cur = x; cur <= n; cur *= x) {
      if (ask(cur)) ans *= x;
    }
  }
  answer(ans);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}