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 maxn = 100000 + 5;
int pri[maxn];
int tot;
int vis[maxn];
int n;
void init(int x) {
  for (int i = 2; i <= x; i++) {
    if (!vis[i]) pri[tot++] = i;
    for (int j = 0; j < tot && i * pri[j] <= x; j++) {
      vis[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  init(n);
  int res = 1, x, sx, p = 0, blk = 69;
  for (int up = sqrt(n); p < tot && pri[p] <= up; p++) {
    int tres = 1;
    if (res * 1ll * pri[p] > 1ll * n) break;
    cout << "B"
         << " " << pri[p] << "\n";
    fflush(stdout);
    cin >> x;
    if (x >= 1) {
      for (int j = pri[p]; j <= n; j *= pri[p]) {
        cout << "A"
             << " " << j << "\n";
        fflush(stdout);
        cin >> sx;
        if (sx >= 1)
          tres = max(tres, j);
        else
          break;
        if (1ll * j * pri[p] > 1ll * n) break;
      }
    }
    res *= tres;
  }
  int pre;
  cout << "A"
       << " " << 1 << "\n";
  fflush(stdout);
  cin >> pre;
  for (int ls = p, rs; ls < tot; ls = rs + 1) {
    rs = min(tot - 1, ls + blk - 1);
    int sum = 0;
    for (int i = ls; i <= rs; i++) {
      cout << "B"
           << " " << pri[i] << "\n";
      fflush(stdout);
      cin >> sx;
      sum += sx;
    }
    cout << "A"
         << " " << 1 << "\n";
    fflush(stdout);
    cin >> sx;
    if ((sx + sum) != pre) {
      for (int i = ls; i <= rs; i++) {
        cout << "A"
             << " " << pri[i] << "\n";
        fflush(stdout);
        cin >> sx;
        if (sx >= 1) {
          res *= pri[i];
          break;
        }
      }
      break;
    } else
      pre = sx;
  }
  cout << "C"
       << " " << res << "\n";
  fflush(stdout);
  return 0;
}
