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>
const long long LL_INF = (long long)2e18 + 5;
using namespace std;
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long N = 1e5, mod = 1e9 + 7;
long long n;
vector<long long> pr(N + 1, true), prime;
long long dividToit;
long long full;
vector<long long> add;
long long ans = 1;
void sieve() {
  pr[0] = false;
  pr[1] = false;
  for (long long i = 2; i * i <= N; i++)
    if (pr[i]) {
      for (long long j = i * i; j <= N; j += i) pr[j] = false;
    }
}
long long query(char ask, long long num) {
  if (num > n) return 0;
  cout << ask << ' ' << num << endl;
  if (ask != 'C') {
    long long ret;
    cin >> ret;
    return ret;
  }
  return -1;
}
long long solve_this(bool there_is_a_fair) {
  long long ct = query('A', 1);
  if (there_is_a_fair or ct != full - add.size()) {
    for (auto i : add) {
      if (query('A', i)) {
        query('C', ans * i);
        return 1;
      }
    }
    return 0;
  }
  full = ct;
  add.clear();
  return 0;
}
void solve() {
  cin >> n;
  vector<long long> b;
  ans = 1;
  for (long long i = (long long)(0); i < (long long)(n + 1); i++)
    if (pr[i]) prime.push_back(i);
  for (auto i : prime) {
    if (i * i <= n) {
      long long res = query('B', i);
      if (res) {
        long long pw = i;
        while (query('A', pw) > 0) {
          ans *= i;
          pw *= i;
        }
      }
    } else
      b.push_back(i);
  }
  dividToit = sqrt((long long)b.size()) + 1;
  full = query('A', 1);
  for (auto i : b) {
    long long before = query('B', i);
    add.push_back(i);
    if (before > 1) {
      if (solve_this(1)) return;
    } else if (add.size() >= dividToit)
      if (solve_this(0)) return;
  }
  if (solve_this(0)) return;
  query('C', ans);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  sieve();
  solve();
}