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 unsigned _mod = 998244353;
unsigned mod;
const long long infi = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long qpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x;
    y >>= 1ll;
    x = x * x;
  }
  return ret;
}
long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % mod;
    y >>= 1ll;
    x = x * x % mod;
  }
  return ret;
}
long long _gcd(long long x, long long y) { return y ? _gcd(y, x % y) : x; }
int a[1000010], n, p[1000010], vis[1000010], cnt, x, f[1000010], num, ans = 1;
int main() {
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    if (!vis[i]) p[++cnt] = i;
    for (int j = 1; j <= cnt; ++j) {
      if (i * p[j] > n) break;
      vis[i * p[j]] = 1;
      if (!(i % p[j])) break;
    }
  }
  for (int i = 1; i <= min(cnt, 66); ++i) {
    cout << "B " << p[i] << endl;
    cin >> x;
    cout << "A " << p[i] << endl;
    cin >> x;
    int tmp = p[i];
    while (x == 1) {
      tmp *= p[i];
      f[++num] = p[i];
      if (tmp > n) break;
      cout << "B " << tmp << endl;
      cin >> x;
      cout << "A " << tmp << endl;
      cin >> x;
    }
  }
  if (num)
    for (int i = 1; i <= num; ++i) ans *= f[i];
  if (ans != 1)
    for (int i = 67; i <= cnt; ++i) {
      cout << "A " << p[i] << endl;
      cin >> x;
      if (x != 1) return cout << "C " << ans * p[i] << endl, 0;
    }
  if (ans != 1) return cout << "C " << ans << endl, 0;
  cout << "A 1" << endl;
  cin >> x;
  int tot = x, k = 0;
  for (int i = 67; i <= cnt; ++i) {
    cout << "B " << p[i] << endl;
    cin >> x;
    k++;
    if (!(k % 100) || i == cnt) {
      cout << "A 1" << endl;
      cin >> x;
      if (x == tot - k)
        tot = x, k = 0;
      else
        for (int j = i - k + 1; j <= i; ++j) {
          cout << "A " << p[j] << endl;
          cin >> x;
          if (x == 1) return cout << "C " << p[j] << endl, 0;
        }
    }
  }
  cout << "C " << ans << endl;
  return 0;
}
