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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 100005;
bool p[N];
bool c[N];
void solv() {
  int n;
  scanf("%d", &n);
  int s = sqrt(n);
  for (int i = 2; i <= n; ++i) p[i] = true;
  for (int i = 2; i <= n; ++i) {
    if (!p[i]) continue;
    for (int j = i + i; j <= n; j += i) p[j] = false;
  }
  int vp = -1;
  for (int i = n; i >= 1; --i) {
    if (p[i]) {
      vp = i;
      break;
    }
  }
  int v0 = -1;
  vector<int> v;
  vector<int> u;
  for (int i = 1; i <= n; ++i) {
    if (!p[i]) continue;
    u.push_back(i);
    printf("B %d\n", i);
    fflush(stdout);
    int q;
    scanf("%d", &q);
    int qq = 0;
    for (int j = i; j <= n; j += i) {
      if (!c[j]) {
        ++qq;
        c[j] = true;
      }
    }
    if (q != qq) {
      v.push_back(i);
    }
    if (v0 == -1) {
      if (((int)(u).size()) == s || i == vp) {
        printf("A 1\n");
        fflush(stdout);
        int q;
        scanf("%d", &q);
        int qq = 0;
        for (int j = 1; j <= n; ++j) {
          if (!c[j]) ++qq;
        }
        if (q != qq) {
          for (int i = 0; i < ((int)(u).size()); ++i) {
            printf("A %d\n", u[i]);
            fflush(stdout);
            int q;
            scanf("%d", &q);
            if (q) {
              v0 = u[i];
              break;
            }
          }
        }
        u.clear();
      }
    }
  }
  if (v0 != -1) v.push_back(v0);
  int ans = 1;
  for (int i = 0; i < ((int)(v).size()); ++i) {
    int q = 1;
    while (1) {
      int x = 1;
      for (int j = 0; j < q + 1; ++j) {
        if (x * 1LL * v[i] > n) {
          x = n + 1;
          break;
        }
        x *= v[i];
      }
      if (x > n) break;
      printf("A %d\n", x);
      fflush(stdout);
      int qq;
      scanf("%d", &qq);
      if (!qq) break;
      ++q;
    }
    for (int j = 0; j < q; ++j) ans *= v[i];
  }
  printf("C %d\n", ans);
  fflush(stdout);
}
int main() {
  solv();
  return 0;
}
