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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
const int N = 1e5 + 10;
bool p[N];
bool checkPrime(int x) { return !p[x]; }
void init() {
  for (int i = 2; i < N; i++) {
    if (p[i]) continue;
    for (int j = i + i; j < N; j += i) {
      p[j] = true;
    }
  }
}
int op = 0;
int n;
int queryA(int x) {
  assert(op++ < 10000);
  if (x > n) return 0;
  cout << "A " << x << endl;
  int ans;
  cin >> ans;
  return ans;
}
bool used[N];
pair<int, int> queryB(int x) {
  assert(op++ < 10000);
  if (x > n) return {0, 0};
  cout << "B " << x << endl;
  int cntDel = 0;
  for (int i = 1; i * x <= n; i++) {
    if (used[i * x]) continue;
    used[i * x] = true;
    cntDel++;
  }
  int ans;
  cin >> ans;
  return {ans, cntDel};
}
int outAns(int x) {
  assert(op++ < 10000);
  cout << "C " << x << endl;
  exit(0);
}
void read() { cin >> n; }
int ans = 1;
void calcSmall(int x) {
  queryB(x);
  int after = queryA(x);
  if (after == 0) return;
  ans *= x;
  int a = x * x;
  while (a <= n && queryA(a)) {
    ans *= x;
    a *= x;
  }
}
int lastI;
void calcSmall() {
  for (int i = 2; i * i <= n; i++) {
    lastI = i;
    if (checkPrime(i)) {
      calcSmall(i);
    }
  }
}
void calcBig() {
  int block = 95;
  vector<int> have;
  for (int i = lastI + 1; i <= n; i++) {
    if (checkPrime(i)) {
      have.push_back(i);
    }
  }
  int cntDel = 0;
  reverse((have).begin(), (have).end());
  for (int i = 0; i < (int)have.size(); i += block) {
    for (int j = 0; i + j < (int)have.size() && j < block; j++) {
      cntDel += queryB(have[i + j]).second;
    }
    int now = queryA(1);
    if (n - cntDel != now) {
      for (int j = 0; i + j < (int)have.size() && j < block; j++) {
        if (queryA(have[i + j])) {
          ans *= have[i + j];
          return;
        }
      }
      assert(false);
    }
  }
}
void run() {
  lastI = 1;
  for (int i = 2; i * i <= n; i++) {
    lastI = i;
  }
  ans = 1;
  calcBig();
  calcSmall();
}
void write() { outAns(ans); }
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(20),
      cout.setf(ios::fixed);
  init();
  read();
  run();
  write();
  return 0;
}
