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>
int Max(int a, int b) { return a > b ? a : b; }
int Min(int a, int b) { return a < b ? a : b; }
int Abs(int a) { return a < 0 ? -a : a; }
long long sqr(int a) { return (long long)a * a; }
int qpow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * a * ans % 1000000007;
    a = 1ll * a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
struct node {
  int t, a, b, id;
  bool operator<(const node &y) const { return t < y.t; }
};
namespace ques {
int flg[100010];
int ans, n;
void setup(int _n, int _ans) { n = _n, ans = _ans; }
int ask(int x) {
  int a = 0;
  for (int i = x; i <= n; i += x) a += !flg[i];
  return a;
}
int del(int x) {
  int a = 0;
  for (int i = x; i <= n; i += x) a += !flg[i], (i != ans) && (flg[i] = 1);
  return a;
}
void check(int x) {
  if (x == ans)
    printf("OHHHHHHHHHHHHHHHHHHHHHH");
  else
    printf("No. ans = %d", ans);
}
}  // namespace ques
int T;
int n;
int pri[100010], pri_cnt, be_pri[100010];
int flg[100010];
int main() {
  T = 1;
  while (T--) {
    scanf("%d", &n);
    for (int i = 2; i <= n; ++i) {
      if (!be_pri[i]) pri[++pri_cnt] = i;
      for (int j = 1; j <= pri_cnt && i * pri[j] <= n; ++j) {
        be_pri[i * pri[j]] = 1;
        if (!(i % pri[j])) break;
      }
    }
    int x;
    int ans = 1, unu = n, flag = 0;
    for (int i = 1; i <= pri_cnt; ++i) {
      printf("B %d\n", pri[i]), fflush(stdout);
      scanf("%d", &x);
      int cnt = 0;
      for (int j = pri[i]; j <= n; j += pri[i])
        if (!flg[j]) flg[j] = 1, ++cnt, --unu;
      if (cnt != x)
        while (x > 0) {
          ans *= pri[i];
          if (1ll * ans * pri[i] > n) break;
          printf("A %d\n", ans * pri[i]), fflush(stdout);
          scanf("%d", &x);
        }
      if ((!(i % (int)sqrt(pri_cnt)) || i == pri_cnt) && !flag) {
        printf("A 1\n"), fflush(stdout);
        scanf("%d", &x);
        if (x != unu)
          for (int j = i - (int)sqrt(pri_cnt) + 1; j <= i; ++j) {
            x = 1;
            while (x) {
              if (1ll * ans * pri[j] > n) break;
              printf("A %d\n", ans * pri[j]), fflush(stdout);
              scanf("%d", &x);
              if (!x && !flag) goto qaq;
              if (!x) break;
              ans *= pri[j];
              flag = 1;
            }
            break;
          qaq:;
          }
      }
    }
    printf("C %d\n", ans), fflush(stdout);
  }
}
