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 int MX = 1e5 + 5;
bitset<MX> val;
long long n, res;
set<long long> primes;
vector<pair<int, int> > pr;
void pre() {
  val.set();
  for (int i = 2; i <= n; ++i) {
    if (val[i]) {
      int cnt = 0;
      for (int j = i; j <= n; j += i) {
        if (val[j]) ++cnt;
        val[j] = 0;
      }
      pr.push_back({i, cnt});
    }
  }
}
bool ok(long long x) {
  cout << "A " << x << endl;
  cin >> res;
  return (res == 1);
}
long long bsearch(long long fac, long long l, long long r) {
  long long lf = 1, rf = 1;
  for (int i = 0; i < (l); ++i) lf *= fac;
  for (int i = 0; i < (r); ++i) rf *= fac;
  if (l == r) return r;
  if (l == r - 1) {
    if (ok(rf)) return r;
    return l;
  } else {
    long long m = (l + r) / 2;
    long long mf = 1;
    for (int i = 0; i < (m); ++i) mf *= fac;
    if (ok(mf))
      return bsearch(fac, m, r);
    else
      return bsearch(fac, l, m - 1);
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  pre();
  if (n == 1) {
    cout << "C 1" << endl;
    return 0;
  }
  long long removed = 0, index = 0;
  bool firstFound = false;
  for (int i = 0; i < ((int)(pr).size()); ++i) {
    cout << "B " << pr[i].first << endl;
    cin >> res;
    removed += res;
    if (res > pr[i].second) {
      primes.insert(pr[i].first);
    }
    ++index;
    if (index == 250 || i + 1 == (int)(pr).size()) {
      cout << "A 1" << endl;
      cin >> res;
      if (res != n - removed && !firstFound) {
        firstFound = true;
        int st = i - index + 1;
        for (int j = st; j < (i + 1); ++j) {
          cout << "A " << pr[j].first << endl;
          cin >> res;
          if (res) {
            primes.insert(pr[j].first);
            break;
          }
        }
      }
      index = 0;
    }
  }
  long long ans = 1;
  for (auto i : primes) {
    long long mx = 1, product = i;
    while (product <= n) {
      ++mx;
      product *= i;
    }
    long long times = bsearch(i, 1, mx - 1);
    for (int j = 0; j < (times); ++j) ans *= i;
  }
  cout << "C " << ans << endl;
}
