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;
long long power(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long pr[100005];
vector<long long> comp;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> v;
  for (long long i = 2; i <= n; i++) {
    if (pr[i] == 0) {
      v.push_back(i);
      for (long long j = i; j <= n; j += i) {
        pr[j] = 1;
      }
    }
  }
  long long sz = v.size();
  long long x = sqrt(sz);
  long long y = 0;
  if (x * x == sz) {
    y = x;
  } else {
    x++;
    y = x;
  }
  long long tot = n;
  for (long long i = 0; i < y; i++) {
    long long mm = 0;
    long long z;
    for (long long j = 0; j < x && i * y + j < sz; j++) {
      cout << "B " << v[i * y + j] << endl;
      cin >> z;
      mm += z;
    }
    cout << "A " << 1 << endl;
    cin >> z;
    if (tot - mm != z) {
      comp.push_back(i);
    }
    tot = z;
  }
  long long ans = 1;
  vector<long long> total;
  long long z;
  for (auto i : comp) {
    for (long long j = 0; j < x && i * y + j < sz; j++) {
      cout << "A " << v[i * y + j] << endl;
      cin >> z;
      if (z) total.push_back(v[i * y + j]);
    }
  }
  for (auto i : total) {
    long long x = i, y = i;
    long long z;
    x *= i;
    while (x <= n) {
      cout << "A " << x << endl;
      cin >> z;
      if (z == 0) break;
      x *= i;
    }
    ans *= (x / i);
  }
  cout << "C " << ans << endl;
}