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;
template <typename A, typename B>
struct unmap {
  struct node {
    A u;
    B v;
    long long nxt;
  };
  static const unsigned uN = 2000003;
  vector<node> e;
  long long head[uN];
  unmap() { memset(head, -1, sizeof(head)); }
  bool count(A u) {
    long long hs = u % uN;
    for (long long i = head[hs]; ~i; i = e[i].nxt)
      if (e[i].u == u) return 1;
    return 0;
  }
  B &operator[](A u) {
    long long hs = u % uN;
    for (long long i = head[hs]; ~i; i = e[i].nxt)
      if (e[i].u == u) return e[i].v;
    e.push_back({u, B(), head[hs]}), head[hs] = e.size() - 1;
    return e.back().v;
  }
  void clear() { memset(head, -1, sizeof(head)); }
};
bool vis[100002];
long long pri[100002], tot, n, x, sum, ans, ia;
signed main() {
  srand(19260817);
  scanf("%lld", &n);
  for (register long long i = 2; i <= n; ++i) {
    if (!vis[i]) {
      pri[++tot] = i;
      if (i <= sqrt(n))
        for (register long long j = i * i; j <= n; j += i) vis[j] = 1;
    }
  }
  memset(vis, 0, sizeof(vis));
  register long long k = sqrt(tot);
  sum = n;
  ans = 1;
  for (register long long i = 1; i <= tot; ++i) {
    if (i >= k && ans * pri[i - k + 1] > n) break;
    printf("B %lld", pri[i]);
    cout << endl;
    register long long num = 0;
    for (register long long j = pri[i]; j <= n; j += pri[i]) {
      if (!vis[j]) {
        ++num;
        --sum;
        vis[j] = 1;
      }
    }
    scanf("%lld", &x);
    if (x != num) {
      for (register long long kk = pri[i]; kk <= n; kk *= pri[i]) {
        printf("A %lld", kk);
        cout << endl;
        scanf("%lld", &x);
        if (x)
          ans *= pri[i];
        else
          break;
      }
    }
    if ((i == tot || i % k == 0) && !ia) {
      printf("A 1");
      cout << endl;
      scanf("%lld", &x);
      if (x != sum) {
        for (register long long j = i - k + 1; j <= i; ++j) {
          for (register long long kk = pri[j]; kk <= n; kk *= pri[j]) {
            printf("A %lld", kk);
            cout << endl;
            scanf("%lld", &x);
            if (x)
              ans *= pri[j], ia = 1;
            else
              break;
          }
          if (ia) break;
        }
      }
    }
  }
  printf("C %lld", ans);
  cout << endl;
}