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 A(int a) {
  cout << "A " << a << endl;
  cout << flush;
  int temp;
  cin >> temp;
  return temp;
}
int B(int a) {
  cout << "B " << a << endl;
  cout << flush;
  int temp;
  cin >> temp;
  return temp;
}
void C(int a) {
  cout << "C " << a << endl;
  return;
}
int N;
int mid;
int midId;
vector<int> primes;
bool compound[100100];
void erat() {
  for (long long i = 2; i <= N; i++) {
    if (compound[i]) continue;
    primes.push_back(i);
    for (long long j = i * i; j <= N; j += i) compound[j] = true;
  }
}
int ALG1() {
  int begin = A(1);
  int start = midId;
  int end = -1;
  int cnt = 0;
  for (long long i = midId; i <= primes.size() - 1; i++) {
    int p = primes[i];
    cnt += B(p);
    if (cnt % 100 == 99) {
      if (A(1) != begin - cnt) {
        end = i;
        start = max(midId, end - 99);
        break;
      }
    }
  }
  if (end == -1) {
    end = primes.size() - 1;
    start = max(midId, end - 99);
  }
  assert(end - start < 101);
  for (long long i = start; i <= end; i++) {
    int p = primes[i];
    if (B(p) == 1) {
      return p;
    }
  }
  return 1;
}
void ALG2() {
  long long prod = 1;
  for (auto p : primes) {
    if (p > mid) break;
    if (prod * p > N) break;
    long long exprime = 1;
    while (1) {
      if (prod * exprime * p > N) break;
      if (B(exprime * p) == 0) break;
      exprime *= p;
    }
    prod *= exprime;
  }
  int other = ALG1();
  C(prod * other);
}
void solve() {
  cin >> N;
  if (N == 1) {
    C(1);
    return;
  }
  erat();
  mid = floor(sqrt((double)N));
  int rmCnt = 0;
  for (long long i = 0; i < primes.size(); i++) {
    int p = primes[i];
    midId = i;
    if (p > mid) break;
    rmCnt += B(p);
  }
  int allCnt = A(1);
  if (allCnt + rmCnt == N)
    C(ALG1());
  else
    ALG2();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
}
