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;
#pragma GCC optimize("O3")
long long i, j, k, l, n, m, ans = 1;
long long a[100009];
vector<long long> p, res;
long long query(long long t, long long x) {
  long long res = 0;
  if (t == 1) {
    cout << 'A' << ' ' << x << '\n';
    cout.flush();
    cin >> res;
  }
  if (t == 2) {
    cout << 'B' << ' ' << x << '\n';
    cout.flush();
    cin >> res;
  }
  if (t == 3) {
    cout << 'C' << ' ' << x << '\n';
    cout.flush();
    exit(0);
  }
  return res;
}
int main() {
  cin >> n;
  p.push_back(0);
  for (i = 2; i <= n; i++) {
    if (a[i] == 0) {
      for (j = i * i; j <= n; j += i) a[j] = 1;
      p.push_back(i);
    }
  }
  m = p.size() - 1;
  long long flag = 0, l = 0, r = 0;
  long long bl = sqrt(m) + 1;
  memset(a, 0, sizeof(a));
  for (i = 1; i <= m; i++) {
    k = query(2, p[i]);
    long long cnt = 0;
    for (j = p[i]; j <= n; j += p[i]) {
      cnt += a[j] == 0;
      a[j] = 1;
    }
    if (cnt != k) {
      res.push_back(p[i]);
    }
    if (flag) continue;
    if (i % bl == 0 || i == m) {
      k = query(1, 1);
      cnt = 0;
      for (j = 1; j <= n; j++) {
        cnt += a[j] == 0;
      }
      if (cnt != k) {
        flag = 1;
        l = ((i - 1) / bl) * bl + 1;
        r = i;
      }
    }
  }
  if (l != 0) {
    for (i = l; i <= r; i++) {
      if (query(1, p[i]) > 0) {
        res.push_back(p[i]);
        break;
      }
    }
  }
  for (auto to : res) {
    long long num = to;
    while (num * to <= n) {
      num *= to;
      if (query(1, num) == 0) {
        num /= to;
        break;
      }
    }
    ans *= num;
  }
  query(3, ans);
  exit(0);
}
