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>
#pragma GCC optimize(O1)
#pragma GCC optimize(O2)
#pragma GCC optimize(O3)
using namespace std;
const int N = 1e5 + 5;
bool isprime[N];
int prime[N], cnt, n;
inline void sieve() {
  memset(isprime, true, sizeof(isprime));
  isprime[1] = false;
  for (register int i = 2; i <= 1e5; i++) {
    if (isprime[i]) prime[++cnt] = i;
    for (register int j = 1; j <= cnt && i * prime[j] <= 1e5; j++) {
      isprime[i * prime[j]] = false;
      if (i % prime[j] == 0) break;
    }
  }
}
inline int A(int x) {
  printf("A %d\n", x);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
inline int B(int x) {
  printf("B %d\n", x);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
inline int power(int x, int y) {
  int base = x, res = 1;
  while (y) {
    if (y & 1) res *= base;
    base *= base;
    y >>= 1;
  }
  return res;
}
inline int log(int x, int y) {
  int res = 0, f = 1;
  while (f <= y) f *= x, res++;
  return res - 1;
}
int main() {
  sieve();
  while (~scanf("%d", &n)) {
    register int x = 1;
    for (register int i = 1; i <= 100; i++) {
      if (prime[i] > n) break;
      B(prime[i]);
      register int l = 1, r = log(prime[i], n), ans = 0;
      while (l <= r) {
        int mid = (l + r) >> 1, cur = power(prime[i], mid);
        int res = A(cur);
        if (res)
          ans = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      x *= power(prime[i], ans);
    }
    if (x > 1) {
      for (register int i = 101; i <= cnt && prime[i] <= n / x; i++) {
        if (A(prime[i] * x)) x *= prime[i];
      }
    } else {
      register int All = A(1);
      for (register int limit = 200; limit <= 9500; limit += 100) {
        for (register int i = limit - 99; i <= limit; i++) {
          if (prime[i] > n) break;
          B(prime[i]);
        }
        if (All - A(1) == 100)
          All -= 100;
        else {
          for (register int i = limit - 99; i <= limit; i++)
            if (prime[i] <= n && A(prime[i])) {
              x = prime[i];
              break;
            }
          break;
        }
      }
      for (register int i = 9501; i <= cnt; i++) {
        if (prime[i] > n) break;
        B(prime[i]);
      }
      if (A(1)) {
        for (register int i = 9501; i <= cnt; i++)
          if (prime[i] <= n && A(prime[i])) {
            x = prime[i];
            break;
          }
      }
    }
    printf("C %d\n", x);
    fflush(stdout);
  }
  return 0;
}
