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;
using ll = long long;
using ld = double;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
ll n, bl;
vector<ll> p;
ll left;
ll get(ll k) {
  assert(k >= 1 && k <= n);
  cout << "A " << k << endl;
  cout.flush();
  ll res;
  cin >> res;
  return res;
}
ll del(ll k) {
  cout << "B " << k << endl;
  cout.flush();
  ll res;
  cin >> res;
  return res;
}
ll output(ll k) {
  cout << "C " << k << endl;
  cout.flush();
  exit(0);
}
void solve() {
  set<ll> st;
  for (ll i = 0; i < (ll)n; ++i) st.insert(i + 1);
  for (ll i = 0; i < (ll)bl; ++i) {
    cerr << "prime " << p[i] << " " << n << endl;
    if (p[i] <= n) {
      if (del(p[i])) {
        for (int j = p[i]; j <= n; j += p[i]) st.erase(j);
      }
    }
  }
  ll expected = get(1);
  bool hasSmallPrime = expected != st.size();
  cerr << "expected if no small prime: " << st.size() << endl;
  if (!hasSmallPrime) {
    cerr << "finding big primes" << endl;
    for (int i = bl; i < p.size(); i += bl) {
      ll dl = 0;
      if (p[i] > n) break;
      for (int j = i; j < i + bl; ++j) {
        if (p[j] > n) break;
        del(p[j]);
        dl++;
      }
      ll got = get(1);
      if (got != expected - dl) {
        for (int j = i; j < i + bl; ++j) {
          if (get(p[j])) {
            output(p[j]);
          }
        }
      }
      expected = got;
    }
    output(1);
  } else {
    cerr << "finding the small div" << endl;
    ll div;
    for (ll i = 0; i < (ll)bl; ++i) {
      if (get(p[i])) {
        div = p[i];
        break;
      }
    }
    bool go = true;
    while (go) {
      go = false;
      for (ll i = 0; i < (ll)p.size(); ++i) {
        if (p[i] * div > n) continue;
        if (get(p[i] * div)) {
          div = p[i] * div;
          go = true;
        }
      }
    }
    output(div);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  p.push_back(2);
  for (int i = 3; i <= 130000; ++i) {
    bool ok = true;
    for (ll pr : p) {
      if (i % pr == 0) {
        ok = false;
        break;
      }
    }
    if (ok) p.push_back(i);
  }
  cin >> n;
  bl = max((int)sqrt(n) + 2, 2);
  cerr << "bl: " << bl << endl;
  solve();
  return 0;
}
