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 double pi = acos(-1.0);
const double eps = 1e-8;
const int mxn = 100033;
const int mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) f = c == '-' ? -1 : 1, c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * f;
}
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long ksm(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long inv2(long long a, long long mod) { return ksm(a, mod - 2, mod); }
void exgcd(long long a, long long b, long long &x, long long &y, long long &d) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, y, x, d);
    y -= x * (a / b);
  }
}
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int n, prime[mxn];
bool isnp[mxn];
void euler() {
  memset((isnp), (0), sizeof((isnp)));
  isnp[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!isnp[i]) prime[++prime[0]] = i;
    for (int j = 1; j <= prime[0] && i * prime[j] <= n; ++j) {
      isnp[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int x;
long long res = 1;
void query(int opt, int v) {
  if (opt == 1) {
    cout << "A"
         << " " << v << endl;
    cin >> x;
  } else if (opt == 2) {
    cout << "B"
         << " " << v << endl;
    cin >> x;
  } else
    cout << "C"
         << " " << res << endl;
}
void find(int p) {
  while (res * p <= n) {
    query(1, res * p);
    if (x)
      res *= p;
    else
      break;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  ;
  cin >> n;
  euler();
  int sz = sqrt(prime[0]);
  for (int i = 1, cur = n, sum; i <= prime[0]; i += sz) {
    sum = 0;
    for (int j = 0; i + j <= prime[0] && j < sz; ++j) {
      query(2, prime[i + j]);
      sum += x;
    }
    query(1, 1);
    int tmp = x;
    if (sum > cur - x) {
      for (int j = 0; i + j <= prime[0] && j < sz; ++j) find(prime[i + j]);
    }
    cur = tmp;
  }
  query(3, 0);
  return 0;
}
