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;
const int MAX_N = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const long long LINF = 1e18;
int n, spf[MAX_N];
vector<int> sm_primes, bg_primes;
void sieve() {
  for (int i = 0; i <= n; i++) spf[i] = i;
  for (int i = 2; i * i <= n; i++)
    if (spf[i] == i)
      for (int j = i + i; j <= n; j += i)
        if (spf[j] == j) spf[j] = i;
  for (int i = 2; i <= n; i++) {
    if (spf[i] == i) {
      if ((long long)i * i <= n)
        sm_primes.push_back(i);
      else
        bg_primes.push_back(i);
    }
  }
}
void solve() {
  cin >> n;
  sieve();
  int ans = 1;
  for (int i : sm_primes) {
    cout << "B " << i << endl;
    int res;
    cin >> res;
    int j = i;
    while (true) {
      if (j > n) break;
      cout << "A " << j << endl;
      cin >> res;
      if (res > 0)
        ans *= i;
      else
        break;
      j *= i;
    }
  }
  int BS = sqrt(bg_primes.size());
  int cnt1;
  cout << "A 1" << endl;
  cin >> cnt1;
  for (int b = 0; b < bg_primes.size(); b += BS) {
    int sum = 0;
    for (int i = b; i < min(b + BS, (int)bg_primes.size()); i++) {
      cout << "B " << bg_primes[i] << endl;
      int res;
      cin >> res;
      sum += res;
    }
    int cnt2;
    cout << "A 1" << endl;
    cin >> cnt2;
    if (cnt1 - cnt2 != sum) {
      for (int i = b; i < min(b + BS, (int)bg_primes.size()); i++) {
        cout << "A " << bg_primes[i] << endl;
        int res;
        cin >> res;
        if (res > 0) ans *= bg_primes[i];
      }
    }
    cnt1 = cnt2;
  }
  cout << "C " << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc;
  tc = 1;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
}