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 md = int(1e9) + 7;
const long long mx = 100100;
int del(int a) {
  cout << "B " << a << endl;
  fflush(stdout);
  int ret;
  cin >> ret;
  return ret;
}
int scan(int a) {
  cout << "A " << a << endl;
  fflush(stdout);
  int ret;
  cin >> ret;
  return ret;
}
void printans(int a) {
  cout << "C " << a << endl;
  fflush(stdout);
}
int main() {
  int n;
  cin >> n;
  if (n <= 5000) {
    for (int i = 2; i <= n; i++) del(i);
    int ans = 1;
    for (int i = 2; i <= n; i++) {
      ans = max(ans, scan(i) ? i : 0);
    }
    printans(ans);
  }
  vector<int> isp(mx, 1), vp;
  for (int i = 2; i < mx; i++) {
    if (isp[i]) {
      vp.push_back(i);
      for (long long j = 1LL * i * i; j < mx; j += i) isp[j] = 0;
    }
  }
  int cnt = 0;
  while (cnt < ((int)(vp).size()) && vp[cnt] * vp[cnt] <= n) cnt++;
  long long ans = 1;
  for (int i = 0; i < cnt; i++) {
    del(vp[i]);
    int k = vp[i], cur = 1;
    while (1) {
      int ret = 0;
      if (k <= n) ret = scan(k);
      if (ret > 0) {
        cur = k;
        k *= vp[i];
      } else {
        ans *= cur;
        break;
      }
    }
  }
  if (ans > 1) {
    for (int i = cnt; i < ((int)(vp).size()); i++) {
      if (1LL * ans * vp[i] > n) break;
      if (scan(ans * vp[i])) {
        ans *= vp[i];
        break;
      }
    }
    printans(ans);
  } else {
    int cc = scan(1);
    int l = cnt, r = l;
    while (r < ((int)(vp).size()) && vp[r] <= n) r++;
    while (l + 1 < r) {
      int mid = (l + r) / 2;
      for (int i = l; i < mid; i++) {
        del(vp[i]);
      }
      int ncc = scan(1);
      if (cc - ncc == mid - l)
        l = mid;
      else {
        r = mid;
        while (l + 1 < r) {
          mid = (l + r) / 2;
          int ecnt = 0;
          for (int i = l; i < mid; i++) {
            ecnt += scan(vp[i]);
          }
          if (ecnt == 0)
            l = mid;
          else
            r = mid;
        }
        break;
      }
      cc = ncc;
    }
    del(vp[l]);
    int exist = scan(vp[l]);
    if (exist)
      printans(vp[l]);
    else
      printans(1);
  }
  return 0;
}
