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 p = 1e9 + 7;
long long qpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % p;
    a = a * a % p;
    n >>= 1;
  }
  return res;
}
int pri[N];
int sqr[N];
int tot = 0;
int n;
int m;
int sum;
void pre(int n) {
  for (int i = 2; i <= n; i++) {
    bool f = 1;
    for (int j = 2; j * j <= i; j++) {
      if (i % j == 0) {
        f = 0;
        break;
      }
    }
    if (f) pri[++tot] = i;
  }
}
int qy1(int x) {
  printf("A %d\n", x);
  fflush(stdout);
  int y;
  scanf("%d", &y);
  return y;
}
int qy2(int x) {
  printf("B %d\n", x);
  fflush(stdout);
  int y;
  scanf("%d", &y);
  return y;
}
int out(int x) {
  printf("C %d\n", x);
  exit(0);
}
void dfs(long long x) {
  for (int i = 1; i < m; i++) {
    for (int j = sqr[i]; j < sqr[i + 1]; j++) {
      if (x * pri[j] > n) break;
      sum -= qy2(x * pri[j]);
    }
    if (qy1(1) != sum) {
      for (int j = sqr[i]; j < sqr[i + 1]; j++) {
        if (x * pri[j] > n) break;
        if (qy1(x * pri[j])) {
          sum++;
          dfs(x * pri[j]);
          return;
        }
      }
    }
  }
  out(x);
}
void work() {
  scanf("%d", &n);
  sum = n;
  pre(n);
  int Sqr = sqrt(tot);
  m = 0;
  int cnt = 0;
  sqr[++m] = 1;
  for (int i = 1; i <= tot; i++) {
    cnt++;
    if (cnt == Sqr) {
      sqr[++m] = i + 1;
      cnt = 0;
    }
  }
  if (cnt == 0)
    m--;
  else
    sqr[++m] = tot - cnt + 1;
  sqr[++m] = tot + 1;
  dfs(1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    work();
  }
}
