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 long long N = 1e5;
const long long M = 1e6 + 5;
const long long inf = 1e18;
const long long eps = 1e-5;
inline long long read() {
  long long s = 0;
  register bool neg = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar()) neg |= (c == '-');
  for (; c >= '0' && c <= '9'; s = s * 10 + (c ^ 48), c = getchar())
    ;
  s = (neg ? -s : s);
  return s;
}
long long a, prime[N + 5], top, ans = 1;
bool vis[N + 5];
inline void init() {
  for (long long i = (2); i <= (N); ++i) {
    if (!vis[i]) prime[++top] = i;
    for (long long j = 1; i * prime[j] <= N && j <= top; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
inline long long ask(char ch, long long n) {
  printf("%c %lld\n", ch, n);
  fflush(stdout);
  long long res;
  scanf("%lld", &res);
  return res;
}
inline void calc(long long l, long long r) {
  for (long long i = (l); i <= (r); ++i) {
    long long tmp = 1;
    while (prime[i] * tmp <= a && ask('A', prime[i] * tmp)) tmp *= prime[i];
    ans *= tmp;
  }
}
inline void calc1(long long l, long long r) {
  for (long long i = (l); i <= (r); ++i) {
    if (ask('A', prime[i])) {
      ans *= prime[i];
      return;
    }
  }
}
signed main() {
  init();
  a = read();
  long long b = ceil(sqrt(a)), i, sum = a;
  for (i = 1; prime[i] <= b;) {
    long long tmp = 0, _i = i;
    for (long long j = 1; j <= 100 && prime[i] <= b; j++, i++)
      tmp += ask('B', prime[i]);
    long long res = ask('A', 1);
    if (sum - res != tmp) calc(_i, i - 1);
    sum = res;
  }
  for (; i <= top && prime[i] <= a;) {
    long long tmp = 0, _i = i;
    for (long long j = 1; j <= 100 && prime[i] <= a && i <= top; j++, i++) {
      tmp += ask('B', prime[i]);
    }
    long long res = ask('A', 1);
    if (sum - res != tmp) {
      calc1(_i, i - 1);
      break;
    }
    sum = res;
  }
  printf("C %lld", ans);
  fflush(stdout);
  return 0;
}
