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 Arg1, typename Arg2>
ostream &operator<<(ostream &out, const pair<Arg1, Arg2> &x) {
  return out << "(" << x.first << "," << x.second << ")";
}
template <typename Arg1>
ostream &operator<<(ostream &out, const vector<Arg1> &a) {
  out << "[";
  for (const auto &x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
ostream &operator<<(ostream &out, const set<Arg1> &a) {
  out << "[";
  for (const auto &x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1, typename Arg2>
ostream &operator<<(ostream &out, const map<Arg1, Arg2> &a) {
  out << "[";
  for (const auto &x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
void __trace__(const string name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << " ] " << endl;
}
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1 &&arg1, Args &&...args) {
  const string name = names.substr(0, names.find(','));
  cerr << name << " : " << arg1 << " | ";
  __trace__(names.substr(1 + (int)name.size()), args...);
}
clock_t time_p = clock();
void time_taken() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const int nax = 1e5 + 1;
bitset<nax> is_prime;
vector<int> primes;
void sieve(int maxn) {
  is_prime.set();
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i < maxn; i++) {
    if (!is_prime[i]) continue;
    primes.push_back(i);
    for (int j = i * 2; j < maxn; j += i) {
      is_prime[j] = false;
    }
  }
}
void solve() {}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int getRand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
int n;
void mark(int x);
int del(int x) {
  cout << "B " << x << endl;
  cout << flush;
  mark(x);
  int res;
  cin >> res;
  return res;
}
int decl(int x) {
  cout << "C " << x << endl;
  cout << flush;
  exit(0);
}
int cnt(int x) {
  cout << "A " << x << endl;
  cout << flush;
  int res;
  cin >> res;
  return res;
}
bitset<nax> marked;
int active;
void mark(int p) {
  for (int i = p; i <= n; i += p) {
    if (marked[i]) continue;
    active--;
    marked[i] = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n;
  if (n == 1) {
    decl(1);
    return 0;
  }
  active = n;
  sieve(n + 1);
  vector<int> small, big;
  for (int x : primes) {
    if ((long long)x * x <= n) {
      small.push_back(x);
    } else {
      big.push_back(x);
    }
  }
  int ans = 1;
  int m = (int)(big).size();
  const int B = sqrt(m);
  int num_blocks = (m + B - 1) / B;
  for (int i = 0; i < num_blocks; i++) {
    for (int j = i * B; j - i * B < B and j < m; j++) {
      del(big[j]);
    }
    if (cnt(1) == active + 1) {
      for (int j = i * B; j - i * B < B and j < m; j++) {
        if (cnt(big[j]) == 1) {
          ans = big[j];
          break;
        }
      }
      break;
    }
  }
  for (int x : small) {
    del(x);
    if (cnt(x) == 1) {
      long long y = x * x;
      while (y <= n and cnt(y) == 1) {
        y *= x;
      }
      ans *= y / x;
    }
  }
  decl(ans);
  time_taken();
  return 0;
}
