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>
const int MAXN = 1e5 + 19;
int n, prime[MAXN], cnt;
bool vist[MAXN], isDel[MAXN];
std::vector<int> divd;
int main() {
  std::scanf("%d", &n);
  if (n == 1) {
    std::puts("C 1");
    return 0;
  }
  for (int i = 2; i <= n; ++i) {
    if (!vist[i]) prime[++cnt] = i;
    for (int j = 1; j <= cnt && i * prime[j] <= n; ++j) {
      vist[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  int b = std::sqrt(cnt), sum = n;
  int last = 1, mf = 0;
  for (int x, i = 1; i <= cnt; ++i) {
    int num = 0;
    for (int j = prime[i]; j <= n; j += prime[i])
      if (!isDel[j]) --sum, ++num, isDel[j] = 1;
    std::printf("B %d\n", prime[i]);
    std::fflush(stdout);
    std::scanf("%d", &x);
    if (x ^ num) divd.push_back(prime[i]);
    if (!mf && (i % b == 0 || i == cnt)) {
      std::puts("A 1");
      std::fflush(stdout);
      std::scanf("%d", &x);
      if (x ^ sum) {
        mf = 1;
        for (int j = last; j <= i; ++j) {
          std::printf("A %d\n", prime[j]);
          std::fflush(stdout);
          std::scanf("%d", &x);
          if (x) {
            divd.push_back(prime[j]);
            break;
          }
        }
      }
      last = i + 1;
    }
  }
  int ans = 1;
  for (int d : divd) {
    ans *= d;
    long long v = (long long)d * d;
    while (1) {
      if (v > n) break;
      int x;
      std::printf("A %lld\n", v);
      std::fflush(stdout);
      std::scanf("%d", &x);
      if (x)
        ans *= d, v *= d;
      else
        break;
    }
  }
  std::printf("C %d\n", ans);
}
