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 longs = long long;
using uint = unsigned;
inline int nextInt() {
  int x = 0, f = 0, ch = getchar();
  while (!isdigit(ch)) ch == '-' && (f = 1), ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
  return f ? -x : x;
}
template <int n>
vector<int> &EulerSieve() {
  static vector<int> prime;
  bitset<n> vis;
  for (int i = 2; i <= n; ++i) {
    if (!vis[i]) prime.push_back(i);
    for (auto &pp : prime) {
      if ((longs)i * pp > n) break;
      vis[i * pp] = true;
      if (i % pp == 0) break;
    }
  }
  return prime;
}
const int N = 100000;
int askA(int x) {
  int tmp;
  cout << "A " << x << endl;
  cin >> tmp;
  return tmp;
}
int askB(int x) {
  int tmp;
  cout << "B " << x << endl;
  cin >> tmp;
  return tmp;
}
void askC(longs x) { cout << "C " << x << endl; }
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int n, sq;
  cin >> n;
  const int step = 100;
  auto &p = EulerSieve<N>();
  int size = p.size();
  sq = sqrt(n);
  longs ans = 1;
  for (int i = 0; i < step; ++i) {
    if (ans * p[i] > n) break;
    int a = 1, pro = p[i];
    askB(p[i]);
    do a = askA(pro);
    while (a && (pro *= p[i]) <= n);
    ans *= pro / p[i];
  }
  if (ans == 1) {
    int now = askA(1);
    int lim = (size - 1) / step + 1;
    for (int i = 1; i <= lim; ++i) {
      int bs = i * step, bd = min(size, bs + step);
      int del = 0;
      for (int j = bs; j < bd; ++j)
        if (p[j] > n)
          break;
        else
          askB(p[j]), ++del;
      if (!del) break;
      auto res = askA(1);
      if (now - res != del) {
        for (int k = bs; k < bd; ++k)
          if (askA(p[k])) {
            ans *= p[k];
            break;
          }
        askC(ans);
        return 0;
      } else
        now = res;
    }
  } else
    for (int j = step; j < size && p[j] * ans <= n; ++j)
      if (askB(p[j] * ans)) {
        askC(p[j] * ans);
        return 0;
      }
  askC(ans);
  return 0;
}
