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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = (int)1e5 + 7;
int n;
int lp[N];
vector<int> primes;
vector<int> small;
vector<int> big;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      primes.push_back(i);
    }
    for (int j = 0;
         j < (int)primes.size() && primes[j] <= lp[i] && primes[j] * i <= n;
         j++) {
      lp[primes[j] * i] = primes[j];
    }
  }
  for (auto &x : primes) {
    if ((long long)x * x <= n) {
      small.push_back(x);
    } else {
      big.push_back(x);
    }
  }
  shuffle(small.begin(), small.end(), rnd);
  shuffle(big.begin(), big.end(), rnd);
  int sol = 1;
  for (auto &x : small) {
    if ((long long)sol * x > n) {
      continue;
    }
    cout << "B " << x << endl;
    int r;
    cin >> r;
    cout << "A " << x << endl;
    cin >> r;
    while (r) {
      sol *= x;
      if ((long long)sol * x > n) {
        break;
      }
      cout << "A " << sol * x << endl;
      cin >> r;
    }
  }
  if (sol > 1) {
    for (auto &x : big) {
      if ((long long)sol * x > n) {
        continue;
      }
      cout << "A " << x << endl;
      int r;
      cin >> r;
      if (r == 2) {
        sol *= x;
        break;
      }
    }
  } else {
    int rad = sqrt((int)big.size());
    while (!big.empty()) {
      vector<int> vec;
      int l = min((int)big.size(), rad);
      for (int i = 0; i < l; i++) {
        vec.push_back(big.back());
        big.pop_back();
      }
      for (auto &x : vec) {
        cout << "B " << x << endl;
        int r;
        cin >> r;
      }
      cout << "A " << 1 << endl;
      int r;
      cin >> r;
      if (r == 1 + (int)big.size()) {
        continue;
      }
      for (auto &x : vec) {
        cout << "A " << x << endl;
        int r;
        cin >> r;
        if (r) {
          sol = x;
          break;
        }
      }
      break;
    }
  }
  cout << "C " << sol << endl;
}
