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 ull = unsigned long long;
using ll = long long;
using ld = long double;
const int mod = 1e9 + 7;
const int inf = INT_MAX;
const int N = 1e5 + 5;
int n, total;
bool there[N];
int count(int x) {
  cout << "A " << x << endl;
  int cnt;
  cin >> cnt;
  return cnt;
}
int remove(int x) {
  cout << "B " << x << endl;
  for (int i = x; i <= n; i += x) {
    if (there[i]) {
      there[i] = false;
      total--;
    }
  }
  int cnt;
  cin >> cnt;
  return cnt;
}
int answer(int x) {
  cout << "C " << x << endl;
  exit(0);
}
bool prime[N + 1];
void sieve() {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (int i = 2; i * i <= N; i++) {
    if (prime[i]) {
      for (int j = i * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  sieve();
  memset(there, true, sizeof(there));
  cin >> n;
  total = n;
  int ans = 1;
  for (int i = 1; i * i <= n; i++) {
    if (prime[i]) {
      remove(i);
    }
  }
  if (count(1) > total) {
    for (int i = 1; i * i <= n; i++) {
      if (prime[i]) {
        int pow = i;
        while (pow <= n and count(pow)) {
          ans *= i;
          pow *= i;
        }
      }
    }
    remove(ans);
    for (int i = floor(sqrt(n) + 1); i <= n && ans * ans < n; i++) {
      if (prime[i]) {
        int cnt = 0;
        for (int j = i; j <= n; j += i) {
          cnt += there[j];
        }
        if (count(i) > cnt) {
          ans *= i;
        }
      }
    }
  } else {
    vector<int> primes;
    for (int i = floor(sqrt(n) + 1); i <= n; i++) {
      if (prime[i]) {
        primes.push_back(i);
      }
    }
    int sq = ceil(sqrt(primes.size()));
    for (int i = 0; i < primes.size(); i += sq) {
      for (int j = i; j < min((int)primes.size(), i + sq); j++) {
        remove(primes[j]);
      }
      if (count(1) > total) {
        for (int j = i; j < min((int)primes.size(), i + sq); j++) {
          if (count(primes[j])) {
            ans = primes[j];
          }
        }
        break;
      }
    }
  }
  answer(ans);
  return 0;
}