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 = 2e5 + 5;
const int M = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 10007;
int prime[N];
bool vis[N];
int cnt, t[N], x;
void init(int n) {
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) prime[++cnt] = i;
    for (int j = 1; j <= cnt && i * prime[j] <= n; j++) {
      vis[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
int Find(int l, int r) {
  for (int i = l; i <= r; i++) {
    if ((printf("A %d\n", prime[i]), fflush(stdout), scanf("%d", &x), x))
      return prime[i];
  }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  init(n);
  int i, tot = 0;
  for (i = 1; i <= cnt && prime[i] * prime[i] <= n; i++) {
    (printf("B %d\n", prime[i]), fflush(stdout), scanf("%d", &x));
    if ((printf("A %d\n", prime[i]), fflush(stdout), scanf("%d", &x), x))
      t[++tot] = prime[i];
  }
  int k = i;
  if (tot) {
    int sum = 1;
    for (int i = 1; i <= tot; i++) {
      sum *= t[i];
    }
    for (int i = 1; i <= tot; i++) {
      while (sum * t[i] <= n && (printf("A %d\n", sum * t[i]), fflush(stdout),
                                 scanf("%d", &x), x)) {
        sum *= t[i];
      }
    }
    for (int i = k; i <= cnt; i++) {
      if ((printf("A %d\n", prime[i]), fflush(stdout), scanf("%d", &x), x) ==
          2) {
        printf("C %d\n", sum * prime[i]);
        exit(0);
      }
    }
    printf("C %d\n", sum);
    exit(0);
  }
  int cur, S, ct = 0;
  for (cur = cnt - i + 2, S = sqrt(cur); i <= cnt; ++i)
    if ((printf("B %d\n", prime[i]), fflush(stdout), scanf("%d", &x)), --cur,
        ++ct == S) {
      if ((printf("A %d\n", 1), fflush(stdout), scanf("%d", &x), x) != 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", &x), x) != 1
                     ? Find(cnt - ct + 1, cnt)
                     : 1),
          fflush(stdout)),
         0;
  return 0;
}
