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>
const int maxn = 1e6 + 7;
int vis[maxn], isp[maxn], A[maxn], tl, n, cnt1, X, opt;
int vis2[maxn];
void sieve(int n) {
  for (int i = 1; i <= n; ++i) {
    vis[i] = isp[i] = vis2[i] = 1;
  }
  isp[1] = 0;
  for (int i = 2; i * i <= n; ++i) {
    if (isp[i])
      for (int j = i * 2; j <= n; j += i) isp[j] = 0;
  }
}
int ask(int x) {
  if (!(1 <= x && x <= n)) return 0;
  opt++;
  assert(opt < 10000);
  printf("A %d\n", x);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
int query_true(int x) {
  int false_answer = ask(x);
  for (int i = x; i <= n; i += x) false_answer -= vis[i];
  return false_answer;
}
void del(int x) {
  if (!(1 <= x && x <= n)) return;
  assert(1 <= x && x <= n);
  assert(x != 1);
  opt++;
  assert(opt < 10000);
  printf("B %d\n", x);
  for (int i = x; i <= n; i += x) {
    cnt1 -= vis[i];
    vis[i] = 0;
  }
  fflush(stdout);
  int res = 0;
  scanf("%d", &res);
}
void delete_first() {
  for (int i = 1; i <= 1000 && i <= n; ++i) {
    if (!isp[i]) continue;
    del(i);
  }
}
int count() { return cnt1; }
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("C 1");
    fflush(stdout);
    return 0;
  }
  cnt1 = n;
  sieve(n);
  delete_first();
  int N = count();
  int M = ask(1);
  int x = 1;
  if (M != N) {
    for (int i = 2; i <= n; ++i) {
      if (!isp[i]) continue;
      int d = query_true(i);
      int j = i;
      if (d) x *= i;
      while (d) {
        if ((long long)j * i > n) break;
        j = j * i;
        if (j > n) break;
        int d1 = query_true(j);
        if (!d1)
          break;
        else
          x *= i;
      }
    }
    printf("C %d", x);
    fflush(stdout);
    return 0;
  } else {
    for (int i = 2; i <= n; ++i) {
      if (vis[i]) A[++tl] = i;
    }
    int sz = std::sqrt(tl) + 1;
    std::stack<int> stk;
    for (int i = 1; i <= tl; ++i) {
      del(A[i]);
      stk.push(A[i]);
      if (i == tl || i % sz == 0) {
        int cnt = ask(1);
        if (cnt > count()) {
          while (!stk.empty()) {
            int u = stk.top();
            stk.pop();
            if (ask(u)) {
              printf("C %d", u);
              fflush(stdout);
              return 0;
            }
          }
          assert(1);
        }
        while (!stk.empty()) stk.pop();
      }
    }
    printf("C %d", 1);
    fflush(stdout);
    return 0;
  }
  return 0;
}
