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;
namespace IO {
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline int read() {
  int x = 0, sign = 0;
  char s =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (!isdigit(s))
    sign |= s == '-',
        s = (p1 == p2 &&
                     (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
                 ? EOF
                 : *p1++);
  while (isdigit(s))
    x = (x << 1) + (x << 3) + (s - '0'),
    s = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return sign ? -x : x;
}
inline void print(int x) {
  if (x > 9) print(x / 10);
  (*O++ = x % 10 + '0');
}
}  // namespace IO
namespace math {
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
long long ksm(long long a, long long b) {
  long long s = 1, m = a;
  while (b) {
    if (b & 1) s = s * m % mod;
    m = m * m % mod, b >>= 1;
  }
  return s;
}
long long inv(long long x) { return ksm(x, mod - 2); }
long long fc[maxn], ifc[maxn];
void init_m(int n) {
  fc[0] = 1;
  for (int i = 1; i <= n; i++) fc[i] = fc[i - 1] * i % mod;
  ifc[n] = inv(fc[n]);
  for (int i = n - 1; i >= 0; i--) ifc[i] = ifc[i + 1] * (i + 1) % mod;
}
long long C(long long n, long long m) {
  return fc[n] * ifc[m] % mod * ifc[n - m] % mod;
}
}  // namespace math
using namespace IO;
const int N = 1e5 + 5;
int n, p[N], pre[N], cnt, pr[N], res, pd[N];
int gt, qu, ori;
void del(int x) {
  for (int j = x; j <= ori; j += x)
    if (!pd[j]) res--, pd[j] = 1;
}
void pa(int x) {
  qu++;
  assert(qu < 1e4);
  cout << "A " << x << endl;
  cin >> gt;
}
void emplace_back(int x) {
  qu++;
  assert(qu < 1e4 && x > 1);
  cout << "B " << x << endl;
  cin >> gt;
}
int find(int st) {
  if (st == 0) return 1;
  int lim = st / 2;
  for (int i = st; i > lim; i--) emplace_back(pr[i]), del(pr[i]);
  pa(1);
  if (gt != res) {
    int ans = 1;
    for (int i = st; i > lim; i--) {
      long long tmp = pr[i];
      while (tmp <= n) {
        pa(tmp);
        if (!gt) {
          tmp /= pr[i];
          break;
        }
        tmp *= pr[i];
      }
      if (tmp > n) tmp /= pr[i];
      if (tmp > 1) ans *= tmp, n /= tmp;
    }
    return ans * find(min(lim, pre[n]));
  } else
    return find(lim);
}
void solve() {
  cin >> n, res = ori = n;
  int cal = 0;
  for (int i = 2; i <= n; i++) {
    if (!p[i]) {
      pr[++cnt] = i;
      for (int j = i + i; j <= n; j += i) p[j] = 1;
    }
    pre[i] = pre[i - 1] + !p[i];
  }
  cout << "C " << find(cnt) << endl, assert(qu < 1e4);
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}
