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 int N = 100005;
int n, p[N], t[N], tot, res, cnt, cur, ct, S;
bool vis[N];
inline void init(const int& n) {
  register int i, j;
  for (i = 2; i <= n; ++i) {
    if (!vis[i]) p[++cnt] = i;
    for (j = 1; j <= cnt && i * p[j] <= n; ++j) {
      vis[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
inline int find(const int& l, const int& r) {
  for (register int i = l; i <= r; ++i)
    if ((printf("A %d\n", p[i]), fflush(stdout), scanf("%d", &res), res))
      return p[i];
}
int main() {
  register int i;
  for (scanf("%d", &n), init(n), i = 1; i <= cnt && p[i] * p[i] <= n; ++i)
    if ((printf("B %d\n", p[i]), fflush(stdout), scanf("%d", &res), res),
        (printf("A %d\n", p[i]), fflush(stdout), scanf("%d", &res), res))
      t[++tot] = p[i];
  int tp = i;
  if (tot) {
    for (cur = i = 1; i <= tot; ++i) cur *= t[i];
    for (i = 1; i <= tot; ++i)
      while (cur * t[i] <= n && (printf("A %d\n", cur * t[i]), fflush(stdout),
                                 scanf("%d", &res), res))
        cur *= t[i];
    for (i = tp; i <= cnt; ++i)
      if ((printf("A %d\n", p[i]), fflush(stdout), scanf("%d", &res), res) == 2)
        return (printf("C %d\n", cur * p[i]), fflush(stdout)), 0;
    return (printf("C %d\n", cur), fflush(stdout)), 0;
  }
  for (cur = cnt - i + 2, S = sqrt(cur); i <= cnt; ++i)
    if ((printf("B %d\n", p[i]), fflush(stdout), scanf("%d", &res), res), --cur,
        ++ct == S) {
      if ((printf("A %d\n", 1), fflush(stdout), scanf("%d", &res), res) != cur)
        return (printf("C %d\n", find(i - S + 1, i)), fflush(stdout)), 0;
      ct = 0;
    }
  return (printf("C %d\n", (printf("A %d\n", 1), fflush(stdout),
                            scanf("%d", &res), res) != 1
                               ? find(cnt - ct + 1, cnt)
                               : 1),
          fflush(stdout)),
         0;
}
