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 N = 1e5 + 5;
int vis[N], used[N];
vector<int> prime;
void getprime(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!vis[i]) {
      prime.push_back(i);
    }
    for (int p : prime) {
      int v = p * i;
      if (v > n) break;
      vis[v] = 1;
      if (i % p == 0) break;
    }
  }
}
int ask(int x) {
  cout << "A " << x << endl;
  int p;
  cin >> p;
  return p;
}
int del(int x) {
  cout << "B " << x << endl;
  int p;
  cin >> p;
  return p;
}
int main() {
  int n;
  cin >> n;
  getprime(n);
  int cnt = prime.size();
  int blk = sqrt(cnt);
  int ans = 1;
  int res = n;
  int siz = cnt - 1;
  int rcd = n + 1;
  for (int i = 0; i <= siz; ++i) {
    int nxt = i + blk - 1;
    nxt = min(nxt, siz);
    for (int j = i; j <= nxt; ++j) {
      int x = prime[j];
      del(x);
      for (int k = x; k <= n; k += x)
        if (!used[k]) used[k] = 1, res--;
    }
    int tp = ask(1);
    if (tp != res) {
      rcd = i;
      break;
    }
    i = nxt;
  }
  for (int i = rcd; i <= siz; ++i) {
    int x = prime[i];
    int res = 0;
    if (ans * x > n) break;
    for (long long j = x; j <= n; j *= x) {
      if (!used[j]) res++;
    }
    for (long long j = x; j <= n; j *= x) {
      int tp = ask(j);
      if (tp != res)
        ans *= x;
      else
        break;
      if (!used[j]) res--;
    }
  }
  cout << "C " << ans << endl;
}
