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;
using ll = long long;
const int mod = 1e9 + 7;
const int N = 1e5 + 5;
vector<int> primes;
int lpf[N + 1];
void sieve(int n) {
  for (int i = 2; i <= n; i++) {
    if (lpf[i] == 0) {
      lpf[i] = i;
      primes.push_back(i);
    }
    for (int j = 0; j < (int)primes.size() && primes[j] <= lpf[i] &&
                    1LL * i * primes[j] <= n;
         ++j)
      lpf[i * primes[j]] = primes[j];
  }
}
signed main() {
  4;
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  auto query = [&](char c, int q) {
    cout << c << " " << q << endl;
    int ret;
    cin >> ret;
    return ret;
  };
  int n;
  cin >> n;
  if (n == 1) {
    cout << "C 1" << endl;
    return 0;
  }
  sieve(n);
  int sq = sqrt(n) + 1;
  int answer = 1;
  vector<int> big;
  for (int p : primes) {
    if (1ll * p * p <= n) {
      query('B', p);
      int pp = p;
      while (pp <= n and query('A', pp) > 0) {
        answer *= p;
        pp *= p;
      }
    } else {
      big.push_back(p);
    }
  }
  int l = 0;
  int r = big.size();
  int cur_count = query('A', 1);
  while (l + 1 < r) {
    int mid = (l + r) / 2;
    int p = 0;
    for (int i = l; i < mid; i++) {
      p += query('B', big[i]);
    }
    int new_count = query('A', 1);
    if (cur_count - p < new_count) {
      r = mid;
      cur_count = new_count;
    } else {
      l = mid;
      cur_count = new_count;
    }
  }
  int p = query('B', big[l]);
  int new_count = query('A', 1);
  if (cur_count - p < new_count) {
    answer *= big[l];
  }
  cout << "C " << answer << endl;
}