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;
long long n, ans = 1, res, vis[100004], maxn;
bool IsPrime[100004], ck;
vector<long long> vec, adj;
void sang() {
  fill(begin(IsPrime), end(IsPrime), true);
  IsPrime[1] = false;
  for (long long d = 2; d * d <= maxn; ++d)
    if (IsPrime[d])
      for (long long q = d * d; q <= maxn; q += d) IsPrime[q] = false;
  for (long long q = 1; q <= maxn; q++) {
    if (IsPrime[q]) {
      vec.push_back(q);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  maxn = n;
  ck = false;
  sang();
  res = n;
  long long numb = vec.size();
  for (long long l = 0, r; l < numb; l = r + 1) {
    long long len = sqrt(numb);
    r = min(l + len, numb - 1);
    for (long long i = l; i <= r; i++) {
      long long k;
      cout << "B"
           << " " << vec[i] << '\n';
      cout.flush();
      cin >> k;
      long long cnt = 0;
      for (long long j = vec[i]; j <= n; j += vec[i]) {
        if (!vis[j]) {
          cnt++;
          res--;
          vis[j] = 1;
        }
      }
      if (ck && cnt != k) adj.push_back(vec[i]);
    }
    if (ck) continue;
    long long k;
    cout << "A"
         << " " << 1 << endl;
    cout.flush();
    cin >> k;
    if (k != res) {
      ck = true;
      for (long long i = l; i <= r; i++) {
        long long p;
        cout << "A"
             << " " << vec[i] << endl;
        cout.flush();
        cin >> p;
        if (p) adj.push_back(vec[i]);
      }
    }
  }
  for (auto v : adj) {
    long long l = v, k;
    while (true) {
      if (l > n) break;
      cout << "A"
           << " " << l << endl;
      cout.flush();
      cin >> k;
      if (k == 1)
        ans *= v;
      else
        break;
      l *= v;
    }
  }
  cout << "C"
       << " " << ans << '\n';
}
