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 = 100000;
bool isPrime[N + 1];
void sieve(int N) {
  for (int i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (int j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
void ask(char c, int x) {
  cout << c << " " << x << endl;
  cout.flush();
}
bool vis[N + 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  sieve(n);
  vector<int> prime;
  for (int i = 2; i <= n; i++) {
    if (isPrime[i]) prime.push_back(i);
  }
  vector<int> ans;
  for (auto v : prime) {
    int cnt = 0;
    for (int j = v; j <= n; j += v) {
      if (vis[j] == 0) {
        cnt++;
        vis[j] = 1;
      }
    }
    ans.push_back(cnt);
  }
  int res = 1;
  vector<int> u;
  int tmp = n, x;
  int count = 0;
  for (int v : prime) {
    if (v > 316)
      break;
    else {
      ask('B', v);
      cin >> x;
      tmp -= x;
      ask('A', v);
      cin >> x;
      if (x == 1) {
        u.push_back(v);
        tmp++;
      }
    }
  }
  vector<int> askk;
  for (int i = 0; i < prime.size(); i++) {
    if (prime[i] < 316) continue;
    count++;
    ask('B', prime[i]);
    cin >> x;
    if (x != ans[i]) {
      u.push_back(prime[i]);
      tmp -= (ans[i] - 1);
    } else {
      tmp -= ans[i];
      askk.push_back(prime[i]);
    }
    if (count % 100 == 99 || i == prime.size() - 1) {
      ask('A', 1);
      cin >> x;
      if (x != tmp) {
        tmp = x;
        for (auto v : askk) {
          ask('A', v);
          cin >> x;
          if (x != 0) {
            res *= v;
            break;
          }
        }
        break;
      }
      askk.clear();
    }
  }
  for (int v : u) {
    int t = v;
    while (1) {
      ask('A', t);
      cin >> x;
      if (x == 1) {
        if ((long long)t * v <= (long long)n)
          t *= v;
        else {
          break;
        }
      } else {
        t /= v;
        break;
      }
    }
    res *= t;
  }
  ask('C', res);
  return 0;
}
