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>
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
  putchar(' ');
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
using namespace std;
int n, ans = 1, l, r, cur;
bool dd[100005], mi;
vector<int> prime;
int GetA(int x) {
  cout << "A " << x << endl;
  cin >> x;
  return x;
}
int GetB(int x) {
  cout << "B " << x << endl;
  cin >> x;
  return x;
}
int Get(int x) {
  int res = 0;
  for (int i = x; i <= n; i += x)
    if (!dd[i]) {
      ++res;
      dd[i] = 1;
    }
  return res;
}
int Find(int x) {
  int res = x;
  while (1ll * res * x <= n) {
    if (GetA(res * x))
      res *= x;
    else
      break;
  }
  return res;
}
int main() {
  srand(time(NULL));
  cin >> n;
  for (int i = (2); i <= (n); ++i) {
    if (dd[i]) continue;
    prime.push_back(i);
    if (1ll * i * i > n) continue;
    for (int j = i * i; j <= n; j += i) dd[j] = 1;
  }
  memset(dd, 0, sizeof dd);
  l = 0, r = sqrt(prime.size()) - 1;
  cur = n;
  for (int i = 0; i < prime.size(); ++i) {
    int u = GetB(prime[i]);
    int cnt = Get(prime[i]);
    cur -= cnt;
    if (cnt != u) {
      ans *= Find(prime[i]);
    }
    if (mi) continue;
    if (i == r) {
      u = GetA(1);
      if (u != cur) {
        for (int j = l; j <= r; ++j) {
          if (GetA(prime[j]) != Get(prime[j])) {
            ans *= Find(prime[j]);
            mi = 1;
            break;
          }
        }
      }
      l = r + 1;
      r = min(r + (int)sqrt(prime.size()), (int)prime.size() - 1);
    }
  }
  cout << "C " << ans;
  return 0;
}
