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 int maxn = 1e5 + 5;
int cnt;
int prime[maxn];
bool judge[maxn];
int sum[maxn], sqn[maxn];
inline void init(int n) {
  for (int i = 2; i <= n; i++) {
    if (!judge[i]) {
      prime[++cnt] = i;
      for (int tol = i; tol <= n; tol += i) {
        if (!judge[tol]) {
          sum[i]++;
          judge[tol] = true;
        }
      }
    }
  }
  for (int i = 2; i <= n; i++) sum[i] += sum[i - 1];
}
inline int num(int p, int c) {
  int ans = 1;
  while (c--) {
    ans *= p;
  }
  return ans;
}
inline int cal(int n, int p) {
  int l = 1, r = 1, x;
  while (n >= p) {
    n /= p;
    r++;
  }
  if (r == 2) return p;
  while (l < r) {
    int mid = (l + r) / 2;
    if (mid > 1) {
      cout << "A " << num(p, mid) << '\n';
      cout.flush();
      cin >> x;
    } else
      x = 1;
    if (x)
      l = mid + 1;
    else
      r = mid;
  }
  return num(p, l - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, x, ans = 1, k, sq, tol, c, flag;
  cin >> n;
  tol = n;
  init(n);
  sq = sqrt(cnt);
  flag = cnt + 1;
  for (k = 0; sqn[k] < cnt;) sqn[++k] = k * sq;
  sqn[k] = cnt;
  for (int i = 0; i < k; i++) {
    for (int j = sqn[i] + 1; j <= sqn[i + 1]; j++) {
      cout << "B " << prime[j] << '\n';
      cout.flush();
      cin >> x;
    }
    tol -= sum[prime[sqn[i + 1]]] - sum[prime[sqn[i]]];
    cout << "A 1" << '\n';
    cout.flush();
    cin >> x;
    if (x == tol)
      continue;
    else {
      for (int j = sqn[i] + 1; j <= sqn[i + 1] && n >= prime[j]; j++) {
        cout << "A " << prime[j] << '\n';
        cout.flush();
        cin >> x;
        if (x) {
          c = cal(n, prime[j]);
          ans *= c;
          n /= c;
        }
        flag = j + 1;
      }
      break;
    }
  }
  for (int i = flag; n >= prime[i] && i <= cnt; i++) {
    cout << "A " << prime[i] * ans << '\n';
    cout.flush();
    cin >> x;
    if (x) {
      c = cal(n, prime[i]);
      ans *= c;
      n /= c;
    }
  }
  cout << "C " << ans << '\n';
  cout.flush();
}
