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;
template <typename T>
vector<T>& operator--(vector<T>& v) {
  for (auto& i : v) --i;
  return v;
}
template <typename T>
vector<T>& operator++(vector<T>& v) {
  for (auto& i : v) ++i;
  return v;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  for (auto& i : v) os << i << ' ';
  return os;
}
template <typename T, typename U>
istream& operator>>(istream& is, pair<T, U>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <typename T, typename U>
pair<T, U> operator-(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <typename T, typename U>
pair<T, U> operator+(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <typename T, typename U>
void umin(T& a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
void umax(T& a, U b) {
  if (a < b) a = b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<bool> isp(n + 1, true);
  isp[0] = isp[1] = false;
  for (int i = 2; i * i < n + 1; ++i) {
    if (isp[i]) {
      for (int j = i * i; j < n + 1; j += i) {
        isp[j] = false;
      }
    }
  }
  vector<int> primes;
  for (int i = 0; i < n + 1; ++i) {
    if (isp[i]) {
      primes.push_back(i);
    }
  }
  int cnt = 0;
  auto ask = [&](int d) {
    if (d > n) {
      return 0;
    }
    ++cnt;
    assert(cnt <= 9999);
    if (d == 1) {
      cout << "A 1" << endl;
    } else {
      cout << "B " << d << endl;
    }
    int x;
    cin >> x;
    return x;
  };
  auto say = [&](int d) {
    cout << "C " << d << endl;
    exit(0);
  };
  int x = 1;
  for (int p : primes) {
    if (p * p > n) break;
    ask(p);
    int cur = p;
    while (ask(cur) != 0) {
      cur *= p;
      x *= p;
    }
  }
  if (x == 1) {
    int l = 0;
    while (l < primes.size() && primes[l] * primes[l] <= n) ++l;
    int r = (int)primes.size() - 1;
    while (l <= r) {
      int have = r - l + 1 + 1;
      int m = (l + r) / 2;
      for (int i = l; i <= m; ++i) {
        ask(primes[i]);
      }
      if (ask(1) == r - m + 1) {
        l = m + 1;
        continue;
      }
      for (int i = l; i <= m; ++i) {
        if (ask(primes[i])) {
          x *= primes[i];
        }
      }
      break;
    }
    say(x);
  } else {
    for (int p : primes) {
      if ((long long)p * p <= n) continue;
      if (ask(p) == 2) {
        x *= p;
      }
    }
    say(x);
  }
  return 0;
}
