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 long long inf = 1e9 + 7;
const long long linf = 1e18 + 1;
template <typename T>
class V : public vector<T> {
 public:
  V(long long sz = 0, T init = T()) : vector<T>(sz, init) {}
};
template <typename T = long long>
T nxt() {
  T x;
  cin >> x;
  return x;
}
template <typename T = long long>
void read(vector<T> &x) {
  for (T &i : x) cin >> i;
}
template <typename T = long long>
void read(V<T> &x) {
  for (T &i : x) cin >> i;
}
template <typename T = long long>
void print(vector<T> &x) {
  for (T v : x) {
    cout << v << " ";
  }
}
template <typename T = long long>
void print(V<T> &x) {
  for (T v : x) {
    cout << v << " ";
  }
}
template <typename T = long long>
void printset(set<T> &x) {
  for (auto v : x) cout << v << " ";
  cout << endl;
}
signed main() {
  ;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n = nxt();
  vector<long long> erat(n + 1, 0);
  vector<long long> prime;
  for (long long i = 2; i <= n; ++i) {
    if (!erat[i]) {
      prime.push_back(i);
      for (long long j = 2 * i; j <= n; j += i) {
        erat[j] = 1;
      }
    }
  }
  long long j = 0;
  vector<long long> ans;
  long long op_count = 0;
  for (; j < prime.size() && prime[j] * prime[j] <= n; ++j) {
    cout << "B " << prime[j] << endl;
    cout.flush();
    long long res;
    cin >> res;
    cout << "A " << prime[j] << endl;
    cin >> res;
    if (res) {
      ans.push_back(prime[j]);
    }
    op_count += 2;
  }
  while (j < prime.size()) {
    long long right = min(j + 100, (long long)prime.size());
    cout << "A " << 1 << endl;
    cout.flush();
    long long cnt;
    cin >> cnt;
    long long sum = 0;
    op_count++;
    for (long long i = j; i < right; ++i) {
      long long v = prime[i];
      cout << "B " << v << endl;
      cout.flush();
      long long res;
      cin >> res;
      op_count++;
      sum += res;
    }
    cout << "A " << 1 << endl;
    op_count++;
    cout.flush();
    long long cnt2;
    cin >> cnt2;
    if (cnt2 + sum != cnt) {
      for (long long i = j; i < right; ++i) {
        cout << "A " << prime[i] << endl;
        op_count++;
        cout.flush();
        long long res;
        cin >> res;
        if (res) {
          ans.push_back(prime[i]);
          break;
        }
      }
      break;
    }
    j = right;
  }
  long long x = 1;
  assert(op_count < 9900);
  for (long long v : ans) {
    long long prom = v;
    while (prom <= n) {
      long long nxt_prom = prom * v;
      if (nxt_prom > n) break;
      cout << "A " << nxt_prom << endl;
      op_count++;
      cout.flush();
      long long res;
      cin >> res;
      if (res == -1)
        assert(0);
      else if (res > 0) {
        prom = nxt_prom;
      } else {
        break;
      }
    }
    x *= prom;
  }
  cout << "C " << x << endl;
  return 0;
}
