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;
const int N = 100000 + 10;
const int B = 100;
inline int read() {
  int x = 0;
  bool f = true;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = false;
    c = getchar();
  }
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return f ? x : -x;
}
int tot, p[N];
bool isnp[N];
inline void filt(int w) {
  for (register int i = 2; i <= w; ++i) {
    if (!isnp[i]) p[++tot] = i;
    for (int j = 1; j <= tot && i * p[j] <= w; ++j) {
      isnp[i * p[j]] = true;
      if (!(i % p[j])) break;
    }
  }
}
int n, Ans, cur;
bool tag[N], flag;
vector<int> fac;
inline int askA(int x) {
  cout << "A " << x << endl;
  fflush(stdout);
  int res;
  cin >> res;
  return res;
}
inline int askB(int x) {
  cout << "B " << x << endl;
  fflush(stdout);
  int res;
  cin >> res;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  filt(n);
  cur = n;
  for (register int i = 1; i <= tot; ++i) {
    int ans = askB(p[i]), res = 0;
    for (int j = p[i]; j <= n; j += p[i]) res += !tag[j], tag[j] = true;
    if (res != ans) fac.push_back(p[i]);
    cur -= res;
    if (flag || (i < tot && i % B)) continue;
    ans = askA(1);
    if (ans != cur) {
      flag = true;
      int l = (i - 1) / B * B;
      for (register int j = l + 1; j <= i; ++j) {
        if (askA(p[j])) {
          fac.push_back(p[j]);
          break;
        }
      }
    }
  }
  Ans = 1;
  for (int i : fac) {
    for (long long x = i; x <= n && askA(x); x *= i) Ans *= i;
  }
  cout << "C " << Ans << endl;
  return 0;
}