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 ll = long long;
const int SQ = 320;
const int TRES = 100;
int queries = 0;
int ask(int t, int a) {
  queries++;
  if (t == 1) cout << "A " << a << '\n' << flush;
  if (t == 2) cout << "B " << a << '\n' << flush;
  int res;
  cin >> res;
  return res;
}
const int MAXN = 1e5 + 10;
int n;
vector<ll> primes;
int sv[MAXN];
void sieve() {
  for (int p = 2; p <= n; p++) {
    if (sv[p]) continue;
    primes.push_back(p);
    for (int q = p; q <= n; q += p) sv[q] = p;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  if (n <= 4000) {
    for (int i = n; i >= 2; i--) {
      ask(2, i);
      if (ask(1, i) != 0) {
        cout << "C " << i << '\n' << flush;
        return 0;
      }
    }
    cout << "C 1" << '\n' << flush;
    return 0;
  }
  sieve();
  ll x = 1;
  vector<ll> bigp;
  for (auto p : primes) {
    if (p > SQ) bigp.push_back(p);
  }
  ll rem = n;
  bool hasbig = false;
  for (int i = 0; i < bigp.size(); i++) {
    rem -= n / bigp[i];
    ask(2, bigp[i]);
    if (i % TRES == TRES - 1 && !hasbig) {
      if (ask(1, 1) != rem) {
        for (int j = i - TRES + 1; j <= i; j++) {
          if (ask(1, bigp[j])) {
            x *= bigp[j];
            hasbig = true;
            break;
          }
        }
      }
    }
  }
  int sz = bigp.size();
  if (!hasbig && ask(1, 1) != rem) {
    for (int j = TRES * (sz / TRES); j < sz; j++) {
      if (ask(1, bigp[j])) {
        x *= bigp[j];
        hasbig = true;
        break;
      }
    }
  }
  for (auto p : primes) {
    if (p > SQ) break;
    ask(2, p);
    ll pot = 1;
    while (p * pot * x <= n && ask(1, p * pot)) pot *= p;
    x *= pot;
  }
  assert(queries <= 9999);
  cout << "C " << x << flush;
}