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;
int main() {
  int n;
  cin >> n;
  auto query = [&](char type, int mul) {
    cout << type << ' ' << mul << '\n';
    fflush(stdout);
    int ret;
    cin >> ret;
    return ret;
  };
  vector<int> p;
  vector<bool> mark(n + 1);
  for (int i = 2; i <= n; ++i) {
    if (!mark[i]) {
      p.emplace_back(i);
    }
    for (int j = 0; j < p.size() && i * p[j] <= n; ++j) {
      mark[i * p[j]] = true;
      if (i % p[j] == 0) {
        break;
      }
    }
  }
  char A = 'A';
  char B = 'B';
  char C = 'C';
  int ans = 1;
  vector<int> small;
  vector<int> big;
  for (int i : p) {
    if (1LL * i * i <= n) {
      small.emplace_back(i);
    } else {
      big.emplace_back(i);
    }
  }
  vector<bool> vis(n + 1);
  int tot = n;
  function<void(int)> del = [&](int x) {
    for (int i = x; i <= n; i += x) {
      if (!vis[i]) {
        tot -= 1;
      }
      vis[i] = true;
    }
  };
  int m = big.size();
  int block = (int)ceil(sqrt(m));
  int last = 0;
  for (int i = 0; i < m; ++i) {
    int tmp = query(B, big[i]);
    del(big[i]);
    if ((i + 1) % block == 0 || i == m - 1) {
      int cnt = query(A, 1);
      if (cnt != tot) {
        for (int j = last; j <= i; ++j) {
          if (query(B, big[j]) > 0) {
            ans = ans * big[j];
            break;
          }
        }
        break;
      }
      last = i + 1;
    }
  }
  for (int v : small) {
    int cur = v;
    int tmp = query(B, v);
    int mul = 1;
    while (cur <= n) {
      int cnt = query(B, cur);
      if (cnt > 0) {
        mul = cur;
      } else {
        break;
      }
      cur = cur * v;
    }
    del(v);
    ans = ans * mul;
  }
  query(C, ans);
  return 0;
}