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;
int n, x, lst, nw, p[100005], st[100005];
long long ans = 1;
bool fl, vs[100005], tg[100005];
void sieve(int n) {
  vs[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!vs[i]) p[++p[0]] = i;
    for (int j = 1; i * p[j] <= n; ++j) {
      vs[i * p[j]] = 1;
      if (!(i % p[j])) break;
    }
  }
}
int main() {
  scanf("%d", &n);
  sieve(n);
  for (int i = 1; i <= p[0]; ++i) {
    if (!st[0] && 1ll * p[i] * p[i] > n) {
      lst = i;
      fl = 1;
      break;
    }
    int cnt = 0;
    for (int j = 1; j * p[i] <= n; ++j)
      if (!tg[j * p[i]]) ++cnt, tg[j * p[i]] = 1;
    printf("B %d\n", p[i]);
    fflush(stdout);
    scanf("%d", &x);
    if (cnt < x) st[++st[0]] = p[i];
    if (!st[0] && 1ll * p[i] * p[i] <= n) {
      printf("A %d\n", p[i]);
      fflush(stdout);
      scanf("%d", &x);
      if (x) st[++st[0]] = p[i];
    }
  }
  if (fl) {
    int blk = sqrt(p[0]);
    printf("A 1\n");
    fflush(stdout);
    scanf("%d", &nw);
    for (int i = lst, L = lst, cnt = 0; i <= p[0]; ++i) {
      ++cnt;
      printf("B %d\n", p[i]);
      fflush(stdout);
      scanf("%d", &x);
      if (i / blk + 1 != (i + 1) / blk + 1 || i == p[0]) {
        printf("A 1\n");
        fflush(stdout);
        scanf("%d", &x);
        if (x > nw - cnt) {
          for (int j = L; j <= i; ++j) {
            printf("A %d\n", p[j]);
            fflush(stdout);
            scanf("%d", &x);
            if (x) {
              ans = p[j];
              break;
            }
          }
        }
        L = i + 1;
        if (ans > 1) break;
      }
    }
    printf("C %lld\n", ans);
    return 0;
  }
  for (int i = 1; i <= st[0]; ++i) {
    ans *= st[i];
    for (int j = 1; ans <= n; ++j, ans *= st[i]) {
      printf("A %d\n", ans);
      fflush(stdout);
      scanf("%d", &x);
      if (!x) break;
    }
    ans /= st[i];
  }
  printf("C %lld\n", ans);
  fflush(stdout);
  return 0;
}
