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;
using pii = pair<int, int>;
using lli = long long int;
using iii = tuple<int, int, int>;
using ulli = unsigned long long int;
using ui = unsigned int;
const int maxn = 2e5;
const int maxm = 1e7;
const int maxl = 18;
const int inf = 0x3f3f3f3f;
const lli INF = 0x3f3f3f3f3f3f3f3f;
const char sp = ' ';
const char dl = '\n';
const int mod = 1e9 + 7;
int ask(char c, lli x) {
  cout << c << sp << x << endl;
  int ans;
  cin >> ans;
  return ans;
}
int main() {
  lli n;
  cin >> n;
  vector<bool> isprime(n + 1, true);
  vector<int> prime;
  isprime[0] = isprime[1] = false;
  for (int i = 2; i <= n; i++)
    if (isprime[i]) {
      prime.push_back(i);
      if ((lli)i * i <= n)
        for (int j = i * i; j <= n; j += i) isprime[j] = false;
    }
  vector<vector<lli>> vec;
  int sqrtn = sqrt(n);
  for (int i = 0; i < (int)prime.size(); i++) {
    if (i % sqrtn == 0) vec.push_back({});
    vec.back().push_back(prime[i]);
  }
  lli ans = 1;
  int cnt = n;
  for (int i = 0; i < (int)vec.size(); i++) {
    int dem = 0;
    for (int j = 0; j < (int)vec[i].size(); j++) {
      if (ans * vec[i][j] > n) break;
      dem += ask('B', vec[i][j]);
    }
    int d = ask('A', 1);
    if (d + dem == cnt) {
      cnt -= dem;
      continue;
    }
    cnt = d;
    for (int j = 0; j < (int)vec[i].size(); j++) {
      lli t = vec[i][j];
      while (ans * vec[i][j] <= n && ask('A', t) > 0) {
        ans *= vec[i][j];
        t *= vec[i][j];
      }
    }
  }
  cout << 'C' << ' ' << ans << endl;
  return 0;
}
