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 maxn = 1e5 + 23;
const int B = 40;
int n;
int askA(int x) {
  printf("A %d\n", x);
  assert(x >= 1 && x <= n);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
int askB(int x) {
  assert(x >= 2 && x <= n);
  printf("B %d\n", x);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
int cnt, vis[maxn], prime[maxn];
void sieve(int n) {
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) prime[cnt++] = i;
    for (int j = 0; j < cnt && i * prime[j] <= n; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  scanf("%d", &n);
  vector<int> factor;
  vector<int> f;
  sieve(n);
  int now = n;
  for (int i = 0; i < cnt; i++) {
    int sub = askB(prime[i]);
    now -= sub;
    if (i % B == B - 1 || i == cnt - 1) {
      int _now = askA(1);
      if (_now != now) factor.push_back(i / B);
      now = _now;
    }
  }
  for (int i : factor) {
    for (int j = 0; j < B && B * i + j < cnt; j++) {
      int cnt = askA(prime[B * i + j]);
      if (cnt) f.push_back(prime[B * i + j]);
    }
  }
  int ans = 1;
  for (int i : f) {
    for (long long j = 1ll * i * i;; j *= i) {
      if (j > n || !askA(j)) {
        ans *= (j / i);
        break;
      }
    }
  }
  printf("C %d\n", ans);
}
