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 maxn = 1e5 + 7;
const int inf = INT_MAX;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
bool p[maxn];
long long cnt[maxn];
vector<long long> prime_large, prime_small;
long long A(long long x) {
  cout << "A " << x << endl;
  long long y;
  cin >> y;
  return y;
}
long long B(long long x) {
  cout << "B " << x << endl;
  long long y;
  cin >> y;
  return y;
}
void C(long long x) { cout << "C " << x << endl; }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    if (p[i] == 0) {
      if (i * i <= n)
        prime_small.push_back(i);
      else
        prime_large.push_back(i);
      for (long long j = i; j <= n; j += i) p[j] = 1;
    }
  }
  long long ans = 1;
  long long tmp = n;
  if (((long long)(prime_large).size()) >= 1) {
    long long block = sqrt(((long long)(prime_large).size()));
    for (long long j = 0; j * block < ((long long)(prime_large).size()); j++) {
      for (long long k = 0;
           k < block && j * block + k < ((long long)(prime_large).size());
           k++) {
        long long now = prime_large[j * block + k];
        B(now);
        tmp -= n / now;
      }
      if (A(1) != tmp) {
        for (long long k = 0;
             k < block && j * block + k < ((long long)(prime_large).size());
             k++) {
          long long now = prime_large[j * block + k];
          if (A(now) == 1) {
            ans *= now;
            break;
          }
        }
        break;
      }
    }
  }
  for (long long i : prime_small) {
    vector<long long> v;
    tmp = i;
    while (tmp <= n) {
      v.push_back(tmp);
      tmp *= i;
    }
    reverse((v).begin(), (v).end());
    for (long long j : v) {
      B(j);
      if (A(j) == 1) {
        ans *= j;
        break;
      }
    }
  }
  C(ans);
}
