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 int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int min(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
const long long int N = 1e6 + 5, INF = 1e18, MOD = 1e9 + 7, M = 6000;
int n;
int b[N];
std::vector<int> v1, v2, vans;
int ask(char c, int num) {
  int ret;
  cout << c << " " << num << endl;
  cin >> ret;
  return ret;
}
void ens() {
  int ans = 1;
  int tp = 1;
  for (auto x : vans) {
    tp = 1;
    for (long long int i = x; i <= n; i *= 1ll * x)
      if (ask('A', i)) tp = i;
    ans *= tp;
  }
  cout << "C " << ans << endl;
  exit(0);
}
int main() {
  cin >> n;
  int sq = sqrt(n);
  for (int i = 2; i <= sq; i++) {
    if (!b[i]) {
      v1.push_back(i);
      for (int j = i + i; j <= n; j += i) b[j] = true;
    }
  }
  for (int i = sq + 1; i <= n; i++)
    if (!b[i]) v2.push_back(i);
  for (auto x : v1) ask('B', x);
  for (auto x : v1) {
    if (ask('A', x)) {
      vans.push_back(x);
    }
  }
  int nn = ask('A', 1), nn2, t = 0;
  while (v2.size()) {
    int m = v2.size() / 2;
    t = v2.size() - m;
    for (int i = m; i < v2.size(); i++)
      if (ask('B', v2[i]) > 1) vans.push_back(v2[i]);
    nn2 = ask('A', 1);
    if (nn - t != nn2)
      for (int i = m; i < v2.size(); i++)
        if (ask('A', v2[i])) {
          vans.push_back(v2[i]);
          ens();
        }
    nn = ask('A', 1);
    v2.resize(v2.size() - t);
  }
  ens();
  return 0;
}
