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 = 500050;
const int inf = 0x3f3f3f3f;
const long long llinf = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0);
clock_t TIME__START, TIME__END;
void program_end() {}
int n, totquery, del[N];
int tot, pri[N], vis[N];
void shaipri(int lim) {
  for (int i = 2; i <= lim; ++i) {
    if (!vis[i]) vis[i] = i, pri[++tot] = i;
    for (int j = 1; j <= lim && i * pri[j] <= lim; ++j) {
      vis[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
inline void solve() {
  scanf("%d", &n);
  shaipri(n);
  int i;
  for (i = 1; pri[i] * pri[i] <= n && i <= tot; ++i)
    ;
  int ans = 1, cnt = 0, bl = 200, totdel = 0, lsti = i;
  for (int j = i; j <= tot; ++j) {
    int p = pri[j];
    cnt++;
    printf("B %d\n", p), fflush(stdout), ++totquery;
    int x;
    scanf("%d", &x);
    for (int k = p; k <= n; k += p)
      if (!del[k]) totdel++, del[k] = 1;
    if (cnt == bl || j == tot) {
      printf("A 1\n"), fflush(stdout), ++totquery;
      scanf("%d", &x);
      if (x != n - totdel) {
        for (int k = lsti; k <= j; ++k) {
          printf("A %d\n", pri[k]), fflush(stdout), ++totquery;
          scanf("%d", &x);
          if (x) {
            ans *= pri[k];
            break;
          }
        }
        break;
      }
      cnt = 0;
      lsti = j;
    }
  }
  for (int j = 1; j < i; ++j) {
    int p = pri[j], x;
    printf("B %d\n", p), fflush(stdout), ++totquery;
    scanf("%d", &x);
    bool flag = false;
    while (p <= n / ans) {
      printf("A %d\n", p), fflush(stdout), ++totquery;
      scanf("%d", &x);
      if (x == 0) {
        flag = true;
        ans *= p / pri[j];
        break;
      }
      p *= pri[j];
    }
    if (!flag) ans *= p / pri[j];
  }
  printf("C %d\n", ans), ++totquery;
  assert(totquery <= 10000);
}
int main() {
  TIME__START = clock();
  int Test = 1;
  while (Test--) {
    solve();
  }
  TIME__END = clock();
  program_end();
  return 0;
}
