Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
const long long N = 1e5 + 10;
vector<long long> pre1, pre2;
long long n;
long long unprime[N];
long long prime[N / 10], tot;
void init() {
  unprime[1] = 1;
  for (long long i = 2; i < N; i++) {
    if (!unprime[i]) {
      prime[++tot] = i;
    }
    for (long long j = 1; j <= tot && i * prime[j] < N; j++) {
      unprime[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
long long A(long long x = 1) {
  if (x > n) return 0;
  printf("A %lld \n", x);
  fflush(stdout);
  long long k;
  scanf("%lld", &k);
  return k;
}
long long B(long long x) {
  if (x > n) return 0;
  printf("B %lld\n", x);
  fflush(stdout);
  long long k;
  scanf("%lld", &k);
  return k;
}
long long C(long long x) {
  printf("C %lld\n", x);
  fflush(stdout);
  exit(0);
}
signed main() {
  init();
  n = read();
  for (long long i = 2; i <= n; i++) {
    if (!unprime[i]) {
      if (i <= 316)
        pre1.push_back(i);
      else
        pre2.push_back(i);
    }
  }
  long long res = 1;
  for (long long x : pre1) {
    B(x);
    for (long long i = x;; i *= x) {
      if (A(i))
        res *= x;
      else
        break;
    }
  }
  long long tmp1 = pre2.size();
  if (res != 1) {
    for (long long i = 0; i < tmp1; i += 100) {
      long long tmp2 = min(tmp1, i + 100);
      for (long long j = i; j < tmp2; j++) {
        if (A(res * pre2[j])) C(res * pre2[j]);
      }
    }
    C(res);
  } else {
    long long cnt = tmp1 + 1;
    for (long long i = 0; i < tmp1; i += 100) {
      long long tmp2 = min(tmp1, i + 100);
      for (long long j = i; j < tmp2; j++) {
        B(pre2[j]);
      }
      cnt -= min(100ll, tmp1 - i);
      if (A(1) != cnt) {
        for (long long j = i; j < tmp2; j++) {
          if (A(pre2[j])) C(pre2[j]);
        }
      }
    }
    C(1);
  }
}