Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFLL = 1e18;
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
int p[MAXN];
void fillSieve() {
  for (long long i = 4; i < MAXN; i += 2) p[i] = 2;
  for (long long i = 3; i * i < MAXN; i += 2)
    if (!p[i])
      for (long long j = i * i; j < MAXN; j += 2 * i) p[j] = i;
}
vector<int> primes;
bool vist[MAXN];
int A(long long x) {
  printf("A %lld\n", x);
  fflush(stdout);
  int y;
  scanf("%d", &y);
  return y;
}
int B(long long x) {
  printf("B %lld\n", x);
  fflush(stdout);
  int y;
  scanf("%d", &y);
  return y;
}
void C(long long x) {
  printf("C %lld\n", x);
  fflush(stdout);
}
int main() {
  fillSieve();
  int n;
  scanf("%d", &n);
  for (int i = 2; i < (int)n + 1; i++)
    if (!p[i]) primes.push_back(i);
  int m = primes.size(), k = sqrt(m);
  bool firstFound = false;
  long long ans = 1;
  int setSize = n;
  for (int i = 0; i < (int)m; i++) {
    int x = B(primes[i]);
    int cnt = 0;
    for (int j = primes[i]; j <= n; j += primes[i])
      if (!vist[j]) {
        cnt++;
        vist[j] = true;
      }
    setSize -= cnt;
    if (x != cnt) {
      for (long long pk = primes[i]; pk <= n; pk *= primes[i]) {
        int x = A(pk);
        if (x)
          ans *= primes[i];
        else
          break;
      }
    }
    if (!firstFound && (i == m - 1 || i % k == k - 1)) {
      int lo = (i / k) * k, hi = min(lo + k, m);
      int x = A(1);
      if (x == setSize) continue;
      for (int j = lo; j < (int)hi; j++) {
        for (long long pk = primes[j]; pk <= n; pk *= primes[j]) {
          int x = A(pk);
          if (x) {
            ans *= primes[j];
            firstFound = true;
          } else
            break;
        }
        if (firstFound) break;
      }
    }
  }
  C(ans);
  return 0;
}