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;
const int mod = 1e9 + 7;
const int KINF = 0x3f3f3f3f;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> vis(n + 1, 1);
  vector<int> small, large;
  for (int i = 2; i * i <= n; i++) {
    if (!vis[i]) continue;
    small.emplace_back(i);
    for (int j = i * i; j <= n; j += i) vis[j] = 0;
  }
  int sq = sqrt(n);
  for (int i = (int)sq + 1; i < (int)n + 1; ++i)
    if (vis[i]) large.emplace_back(i);
  for (int& x : small) {
    cout << "B " << x << endl;
    int tmp;
    cin >> tmp;
  }
  int ans = 1;
  for (int& x : small) {
    int c = 1;
    int ret = 1;
    while (ret > 0) {
      c *= x;
      if (c > n)
        ret = 0;
      else {
        cout << "A " << c << endl;
        cin >> ret;
      }
    }
    ans *= c / x;
  }
  cout << "A " << 1 << endl;
  int res;
  cin >> res;
  while (!large.empty()) {
    int sz = large.size();
    int m = sz / 2;
    for (int i = (int)m; i < (int)sz; ++i) {
      cout << "B " << large[i] << endl;
      int tmp;
      cin >> tmp;
      if (tmp > 1) {
        ans *= large[i];
        cout << "C " << ans << endl;
        return 0;
      }
    }
    cout << "A " << 1 << endl;
    int remin;
    cin >> remin;
    if (res - remin != sz - m) {
      for (int i = (int)m; i < (int)sz; ++i) {
        cout << "A " << large[i] << endl;
        int tmp;
        cin >> tmp;
        if (tmp > 0) {
          cout << "C " << large[i] << endl;
          return 0;
        }
      }
    }
    res = remin;
    large.resize(m);
  }
  cout << "C " << ans << endl;
  return 0;
}