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;
const long long N = 1000009, OO = 1000000000;
long long n, p[N], power[20], used[N], checked[N];
void Solve() {
  scanf("%lld", &n);
  long long i, j, cnt, ans = 1, x, ok = 0, k;
  vector<long long> prime;
  for (i = 2; i <= n; ++i) p[i] = 1;
  for (i = 2; i <= n; ++i)
    if (p[i])
      for (j = i * i; j <= n; j += i) p[j] = 0;
  prime.push_back(0);
  for (i = 1; i <= n; ++i) {
    if (p[i]) prime.push_back(i);
  }
  long long K = sqrt(prime.size() - 1), num = 0, sum = n;
  memset(used, 0, sizeof(used));
  for (i = 1; i < prime.size(); ++i) {
    if (i > K && ans * prime[i] > n) break;
    num = 0;
    for (j = prime[i]; j <= n; j += prime[i])
      if (!used[j]) {
        used[j] = 1, ++num, --sum;
      }
    printf("B %lld\n", prime[i]);
    fflush(stdout);
    scanf("%lld", &cnt);
    if (cnt != num) {
      x = 1;
      for (k = 1; k <= 17; ++k) {
        x *= prime[i];
        if (x > n) break;
        printf("A %lld\n", x);
        fflush(stdout);
        scanf("%lld", &cnt);
        if (cnt)
          ans *= prime[i];
        else
          break;
      }
    }
    if ((i % K == 0 || i == prime.size() - 1) && !ok) {
      printf("A 1\n");
      fflush(stdout);
      scanf("%lld", &cnt);
      if (cnt != sum) {
        for (j = i - K + 1; j <= i; ++j) {
          x = 1;
          for (k = 1; k <= 17; ++k) {
            x *= prime[j];
            if (x > n) break;
            printf("A %lld\n", x);
            fflush(stdout);
            scanf("%lld", &cnt);
            if (cnt)
              ans *= prime[j], ok = 1;
            else
              break;
          }
          if (ok) break;
        }
      }
    }
  }
  printf("C %lld\n", ans);
  fflush(stdout);
}
int main() {
  Solve();
  return 0;
}
